1

Unit 1 43 topics

Limits and Continuity

Estimating Limits from Graphs, The Algebra of Limits, Limits of Functions, and more.

1.1 Estimating Limits from Graphs

1. 1.1.1

   The Finite Limit of a Function

   $\lim_{x \to a} f(x) = L$ means $f(x)$ gets as close as you like to $L$ as $x$ gets close to $a$ from both sides. The value $f(a)$ does not matter and may not even exist.

   e.g. Graph with an open circle at $(2, 3)$ and a filled dot at $(2, 5)$: $\lim_{x \to 2} f(x) = 3$, but $f(2) = 5$

   $\displaystyle \lim_{x \to a} f(x) = L$
2. 1.1.2

   The Left and Right-Sided Limits of a Function

   The left-hand limit uses $x$-values just less than $a$. The right-hand limit uses $x$-values just greater than $a$.

   e.g. Graph approaches height 3 from the left of $x = 2$ and height 5 from the right: $\lim_{x \to 2^-} f(x) = 3$, $\lim_{x \to 2^+} f(x) = 5$

   $\displaystyle \lim_{x \to a^-} f(x) \quad \text{(from the left)}$

   $\displaystyle \lim_{x \to a^+} f(x) \quad \text{(from the right)}$
3. 1.1.3

   Finding the Existence of a Limit Using One-Sided Limits

   The two-sided limit exists only when both one-sided limits exist and are equal. If they differ, the limit does not exist (DNE).

   e.g. Left limit 3, right limit 5: $\lim_{x \to 2} f(x)$ DNE

   $\displaystyle \lim_{x \to a} f(x) = L \iff$

   $\displaystyle \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$
4. 1.1.4

   Limits at Infinity from Graphs

   Read the height the graph levels off toward as $x$ moves far right ($x \to \infty$) or far left ($x \to -\infty$). If that height is $L$, the line $y = L$ is a horizontal asymptote. The two ends can have different limits.

   e.g. $f(x) = \tfrac1x + 2$ levels off at 2: $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = 2$

   $\displaystyle \lim_{x \to \pm\infty} f(x) = L$
5. 1.1.5

   Infinite Limits from Graphs

   If $f(x)$ grows without bound as $x \to a$, write $\lim_{x \to a} f(x) = \infty$ (or $-\infty$). The graph has a vertical asymptote at $x = a$. The limit still does not equal a real number.

   e.g. $\lim_{x \to 0} \tfrac{1}{x^2} = \infty$; $\lim_{x \to 0^+} \tfrac1x = \infty$ but $\lim_{x \to 0^-} \tfrac1x = -\infty$

   $\displaystyle \lim_{x \to a} f(x) = \pm\infty$

1.2 The Algebra of Limits

1. 1.2.1

   Limits of Power Functions, and the Constant Rule for Limits

   The limit of a constant is that constant. For a power of $x$, substitute $a$. A constant multiple can be pulled out of a limit.

   e.g. $\lim_{x \to 3} 2x^2 = 2 \cdot 9 = 18$

   $\displaystyle \lim_{x \to a} c = c$

   $\displaystyle \lim_{x \to a} x^n = a^n$

   $\displaystyle \lim_{x \to a} c f(x) = c \lim_{x \to a} f(x)$
2. 1.2.2

   The Sum Rule for Limits

   If both limits exist, the limit of a sum or difference is the sum or difference of the limits.

   e.g. $\lim_{x \to 2} (x^3 + 4x) = 8 + 8 = 16$

   $\displaystyle \lim_{x \to a} [f(x) \pm g(x)]$

   $\displaystyle = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$
3. 1.2.3

   The Product and Quotient Rules for Limits

   If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, the limit of a product is $LM$. The limit of a quotient is $\tfrac{L}{M}$, provided $M \ne 0$.

   e.g. $\lim_{x \to 1} \tfrac{x + 3}{x^2 + 1} = \tfrac{4}{2} = 2$

   $\displaystyle \lim_{x \to a} [f(x)\,g(x)] = LM$

   $\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, \; M \ne 0$
4. 1.2.4

   The Power and Root Rules for Limits

   If $\lim_{x \to a} f(x) = L$, raise $L$ to the power or take its root. For an even root, $L$ must be positive.

   e.g. $\lim_{x \to 1} (2x + 1)^3 = 27$; $\lim_{x \to 4} \sqrt{x + 5} = 3$

   $\displaystyle \lim_{x \to a} [f(x)]^n = L^n$

   $\displaystyle \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$

1.3 Limits of Functions

1. 1.3.1

   Limits at Infinity of Polynomials

   As $x \to \pm\infty$, the leading term dominates. Its sign and degree decide whether the limit is $\infty$ or $-\infty$. A nonconstant polynomial never has a finite limit at infinity.

   e.g. $\lim_{x \to -\infty} (-2x^3 + x)$: $x^3 \to -\infty$, times $-2$ gives $\infty$

   $\displaystyle \lim_{x \to \pm\infty} p(x)$

   $\displaystyle = \lim_{x \to \pm\infty} a_n x^n$
2. 1.3.2

   Limits of Reciprocal Functions

   For $n > 0$, $\tfrac{1}{x^n} \to 0$ as $x \to \pm\infty$. Near $x = 0$ it blows up: $\tfrac1x$ goes to $\infty$ from the right and $-\infty$ from the left, while $\tfrac{1}{x^2}$ goes to $\infty$ from both sides.

   e.g. $\lim_{x \to \infty} \tfrac{5}{x^2} = 0$; $\lim_{x \to 3^+} \tfrac{1}{x - 3} = \infty$

   $\displaystyle \lim_{x \to \pm\infty} \frac{1}{x^n} = 0$

   $\displaystyle \lim_{x \to 0^+} \frac1x = \infty$

   $\displaystyle \lim_{x \to 0^-} \frac1x = -\infty$
3. 1.3.3

   Limits of Exponential Functions

   Exponentials are continuous, so substitute at a finite point. For $b > 1$, $b^x \to \infty$ as $x \to \infty$ and $b^x \to 0$ as $x \to -\infty$. For $0 < b < 1$ it is the reverse.

   e.g. $\lim_{x \to \infty} \left(\tfrac12\right)^x = 0$; $\lim_{x \to -\infty} (3 + 2^x) = 3$

   $\displaystyle \lim_{x \to \infty} e^x = \infty$

   $\displaystyle \lim_{x \to -\infty} e^x = 0$

   $\displaystyle \lim_{x \to \infty} e^{-x} = 0$
4. 1.3.4

   Limits of Logarithmic Functions

   Logs are continuous on $x > 0$, so substitute there. $\ln x \to \infty$ as $x \to \infty$ (slowly), and $\ln x \to -\infty$ as $x \to 0^+$. Only the right-hand limit at 0 makes sense.

   e.g. $\lim_{x \to e} \ln x = 1$; $\lim_{x \to 2^+} \ln(x - 2) = -\infty$

   $\displaystyle \lim_{x \to \infty} \ln x = \infty$

   $\displaystyle \lim_{x \to 0^+} \ln x = -\infty$
5. 1.3.5

   Limits of Radical Functions

   Roots are continuous on their domains, so substitute. Even roots need the inside $\ge 0$; at an endpoint of the domain, use the one-sided limit from inside the domain.

   e.g. $\lim_{x \to 5} \sqrt{x + 4} = \sqrt9 = 3$

   $\displaystyle \lim_{x \to 0^+} \sqrt{x} = 0$

   $\displaystyle \lim_{x \to \infty} \sqrt{x} = \infty$

   $\displaystyle \lim_{x \to -\infty} \sqrt[3]{x} = -\infty$
6. 1.3.6

   Limits of Trigonometric Functions

   $\sin x$ and $\cos x$ are continuous everywhere, so substitute. $\tan x$ is continuous except at odd multiples of $\tfrac{\pi}{2}$, where it has vertical asymptotes. $\sin x$ and $\cos x$ have no limit as $x \to \pm\infty$ because they keep oscillating.

   e.g. $\lim_{x \to \pi/3} \cos x = \tfrac12$

   $\displaystyle \lim_{x \to a} \sin x = \sin a$

   $\displaystyle \lim_{x \to a} \cos x = \cos a$

   $\displaystyle \lim_{x \to (\pi/2)^-} \tan x = \infty$
7. 1.3.7

   Limits of Reciprocal Trigonometric Functions

   Write $\csc$, $\sec$, and $\cot$ as reciprocals and substitute where the denominator is not 0. Where the denominator is 0 there is a vertical asymptote; check the sign on each side.

   e.g. $\lim_{x \to \pi/3} \sec x = \tfrac{1}{1/2} = 2$; $\lim_{x \to (\pi/2)^-} \sec x = \infty$

   $\displaystyle \csc x = \frac{1}{\sin x}, \; \sec x = \frac{1}{\cos x}$

   $\displaystyle \lim_{x \to 0^+} \csc x = \infty$

   $\displaystyle \lim_{x \to 0^-} \csc x = -\infty$
8. 1.3.8

   Limits of Piecewise Functions

   Away from a boundary, use the piece that applies there. At a boundary, find each one-sided limit with its own piece. The limit exists only if they match.

   e.g. Left: $2^2 = 4$; right: $3(2) - 2 = 4$; so $\lim_{x \to 2} f(x) = 4$

   $\displaystyle f(x) = \begin{cases} x^2 & x < 2 \\ 3x - 2 & x \ge 2 \end{cases}$

1.4 Determining Limits Using Algebraic Manipulation

1. 1.4.1

   Calculating Limits of Rational Functions by Factoring

   If substitution gives $\tfrac00$, factor the top and bottom, cancel the common factor, then substitute again.

   e.g. $\lim_{x \to 3} \tfrac{x^2 - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$

2. 1.4.2

   Limits of Absolute Value Functions

   Rewrite the absolute value as two cases, split where the inside is 0. Find the one-sided limits using the correct case on each side.

   e.g. $\tfrac{\lvert x \rvert}{x}$: right limit 1, left limit $-1$, so $\lim_{x \to 0} \tfrac{\lvert x \rvert}{x}$ DNE

   $\displaystyle \lvert x \rvert = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$
3. 1.4.3

   Calculating Limits of Radical Functions Using Conjugate Multiplication

   If a radical expression gives $\tfrac00$, multiply the top and bottom by the conjugate. Simplify, cancel, then substitute.

   e.g. $\lim_{x \to 4} \tfrac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4} \tfrac{1}{\sqrt{x} + 2} = \tfrac14$

   $\displaystyle (\sqrt{a} - b)(\sqrt{a} + b) = a - b^2$
4. 1.4.4

   Calculating Limits Using Trigonometric Identities

   If substitution gives $\tfrac00$, rewrite with identities (Pythagorean, double-angle, $\tan x = \tfrac{\sin x}{\cos x}$) until a factor cancels, then substitute.

   e.g. $\lim_{x \to 0} \tfrac{\sin 2x}{\sin x} = \lim_{x \to 0} 2\cos x = 2$

   $\displaystyle \sin^2 x + \cos^2 x = 1$

   $\displaystyle \sin 2x = 2 \sin x \cos x$
5. 1.4.5

   Limits at Infinity and Horizontal Asymptotes of Rational Functions

   Compare the degree $n$ of the top with the degree $m$ of the bottom. If $n < m$ the limit is 0. If $n = m$ it is the ratio of leading coefficients. If $n > m$ it is $\pm\infty$ and there is no horizontal asymptote.

   e.g. $\lim_{x \to \infty} \tfrac{3x^2 + 1}{2x^2 - x} = \tfrac32$, so the HA is $y = \tfrac32$

   $\displaystyle n < m: \; L = 0$

   $\displaystyle n = m: \; L = \frac{a_n}{b_m}$

   $\displaystyle n > m: \; L = \pm\infty$
6. 1.4.6

   Evaluating Limits at Infinity by Comparing Relative Magnitudes of Functions

   As $x \to \infty$, logs grow slower than powers, and powers grow slower than exponentials. In a ratio, the faster-growing function wins.

   e.g. $\lim_{x \to \infty} \tfrac{x^3}{e^x} = 0$; $\lim_{x \to \infty} \tfrac{2^x}{x^{10}} = \infty$

   $\displaystyle \ln x \ll x^p \ll b^x$

   $\displaystyle (p > 0, \; b > 1)$
7. 1.4.7

   Evaluating Limits at Infinity of Radical Functions

   Divide the top and bottom by the highest power of $x$. Inside a square root, $\sqrt{x^2} = \lvert x \rvert$, so when $x \to -\infty$, use $\sqrt{x^2} = -x$ and watch the sign.

   e.g. $\lim_{x \to \infty} \tfrac{\sqrt{4x^2 + 1}}{x + 3} = 2$; $\lim_{x \to -\infty} \tfrac{\sqrt{4x^2 + 1}}{x + 3} = -2$

   $\displaystyle \sqrt{x^2} = \lvert x \rvert$

   $\displaystyle x < 0: \; \sqrt{x^2} = -x$
8. 1.4.8

   Vertical Asymptotes of Rational Functions

   Cancel common factors first. Each $x$-value that still makes the denominator zero is a vertical asymptote.

   e.g. $f(x) = \frac{x + 1}{(x - 3)(x + 1)}$: vertical asymptote $x = 3$ only

9. 1.4.9

   Connecting Infinite Limits and Vertical Asymptotes of Rational Functions

   After canceling common factors, a value that makes the denominator 0 (but not the numerator) is a vertical asymptote, and the one-sided limits there are $\pm\infty$. Test a value on each side to get the sign. A factor that cancels gives a hole instead.

   e.g. $f(x) = \tfrac{x + 1}{x - 2}$: $\lim_{x \to 2^+} f(x) = \infty$, $\lim_{x \to 2^-} f(x) = -\infty$, VA $x = 2$

1.5 Special Limits

1. 1.5.1

   The Squeeze Theorem

   If $f$ is trapped between two functions near $a$ and both outer functions have the same limit $L$ at $a$, then $f$ has limit $L$ too. Often used with $-1 \le \sin x \le 1$.

   e.g. $-x^2 \le x^2\sin\tfrac1x \le x^2$ and both bounds go to $0$, so $\lim_{x\to0} x^2\sin\tfrac1x = 0$

   $\displaystyle g(x) \le f(x) \le h(x) \text{ near } a$

   $\displaystyle \lim_{x\to a} g = \lim_{x\to a} h = L$

   $\displaystyle \Rightarrow\; \lim_{x\to a} f(x) = L$
2. 1.5.2

   Special Limits Involving Sine

   As $x \to 0$, $\sin x$ and $x$ become nearly equal, so their ratio goes to $1$. Only true with $x$ in radians.

   e.g. $\lim_{x\to0}\frac{5\sin x}{2x} = \tfrac52 \cdot 1 = \tfrac52$

   $\displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1$

   $\displaystyle \lim_{x\to0} \frac{x}{\sin x} = 1$
3. 1.5.3

   Evaluating Special Limits Involving Sine Using a Substitution

   Let $u = kx$. As $x \to 0$, $u \to 0$, so the sine limit still applies. Make the angle and the denominator match, then fix the constant.

   e.g. $\lim_{x\to0}\frac{\sin 3x}{x}$: let $u = 3x$, then $\lim_{u\to0}\frac{3\sin u}{u} = 3$

   $\displaystyle \lim_{x\to0} \frac{\sin kx}{kx} = 1$

   $\displaystyle \lim_{x\to0} \frac{\sin kx}{x} = k$
4. 1.5.4

   Special Limits Involving Cosine

   As $x \to 0$, $1 - \cos x$ shrinks faster than $x$, so the ratio goes to $0$. Only true with $x$ in radians.

   e.g. $\lim_{x\to0}\frac{\cos x - 1}{3x} = -\tfrac13 \cdot 0 = 0$

   $\displaystyle \lim_{x\to0} \frac{1 - \cos x}{x} = 0$

   $\displaystyle \lim_{x\to0} \frac{1 - \cos x}{x^2} = \frac12$

1.6 Continuity

1. 1.6.1

   Determining Continuity from Graphs

   A function is continuous at a point if you can trace the graph through it without lifting your pencil. Holes, jumps, and vertical asymptotes are discontinuities.

   e.g. Open circle at $(2, 3)$ and a filled dot at $(2, 5)$: not continuous at $x = 2$

2. 1.6.2

   Defining Continuity at a Point

   $f$ is continuous at $x = a$ only if all three conditions hold. If any one fails, $f$ is discontinuous at $a$.

   e.g. $f(x) = x^2 + 1$ at $x = 2$: $f(2) = 5$ and the limit is $5$, so continuous

   $\displaystyle 1.\; f(a) \text{ is defined}$

   $\displaystyle 2.\; \lim_{x\to a} f(x) \text{ exists}$

   $\displaystyle 3.\; \lim_{x\to a} f(x) = f(a)$
3. 1.6.3

   Left and Right Continuity

   $f$ is continuous from the left at $a$ if the left-hand limit equals $f(a)$, and from the right if the right-hand limit equals $f(a)$. Continuous at $a$ means both.

   e.g. $\lfloor x \rfloor$ at $x = 1$: right limit $1 = f(1)$, left limit $0$, so continuous from the right only

   $\displaystyle \text{left: } \lim_{x\to a^-} f(x) = f(a)$

   $\displaystyle \text{right: } \lim_{x\to a^+} f(x) = f(a)$
4. 1.6.4

   Further Continuity of Piecewise Functions

   Each piece is usually continuous on its own interval, so check only the break points. Both pieces must reach the same value there, and it must equal $f(a)$.

   e.g. $f(x) = x + 1$ for $x < 1$, $3x - 1$ for $x \ge 1$: both pieces give $2$ at $x = 1$, so continuous

   $\displaystyle \lim_{x\to a^-} f = \lim_{x\to a^+} f = f(a)$
5. 1.6.5

   Point Discontinuities

   Also called removable discontinuities. The limit exists, but $f(a)$ is undefined or does not equal it. The graph shows a hole.

   e.g. $f(x) = \frac{x^2 - 4}{x - 2}$ has a hole at $(2, 4)$

   $\displaystyle \lim_{x\to a} f(x) = L \ne f(a)$
6. 1.6.6

   Jump Discontinuities

   Both one-sided limits exist but are different, so the graph jumps. The two-sided limit does not exist.

   e.g. $f(x) = \frac{\lvert x \rvert}{x}$ jumps from $-1$ to $1$ at $x = 0$

   $\displaystyle \lim_{x\to a^-} f(x) \ne \lim_{x\to a^+} f(x)$
7. 1.6.7

   Discontinuities Due to Vertical Asymptotes

   Also called infinite discontinuities. At least one one-sided limit is $\infty$ or $-\infty$, so the function is not continuous there.

   e.g. $f(x) = \frac{1}{x - 3}$: $\lim_{x\to3^+} f(x) = \infty$, so infinite discontinuity at $x = 3$

8. 1.6.8

   Continuity Over an Interval

   $f$ is continuous on $(a, b)$ if it is continuous at every point inside. On $[a, b]$ it must also be continuous from the right at $a$ and from the left at $b$.

   e.g. $\sqrt{x}$ is continuous on $[0, \infty)$. $\tfrac1x$ is continuous on $(0, 5]$ but not on $[-1, 1]$

   $\displaystyle \lim_{x\to a^+} f(x) = f(a)$

   $\displaystyle \lim_{x\to b^-} f(x) = f(b)$
9. 1.6.9

   Continuity of Functions

   Polynomial, rational, root, exponential, log, and trig functions are continuous on their domains. Sums, products, compositions, and quotients (where the denominator is not $0$) of continuous functions are continuous.

   e.g. $\frac{x + 1}{x^2 - 9}$ is continuous everywhere except $x = \pm 3$

10. 1.6.10

    The Intermediate Value Theorem

    If $f$ is continuous on $[a, b]$ and $N$ is between $f(a)$ and $f(b)$, then $f$ hits $N$ somewhere in between. Continuity is required.

    e.g. $f(x) = x^3 + x - 1$: $f(0) = -1$, $f(1) = 1$, so $f$ has a root in $(0, 1)$

    $\displaystyle f \text{ continuous on } [a, b]$

    $\displaystyle N \text{ between } f(a) \text{ and } f(b)$

    $\displaystyle \Rightarrow\; f(c) = N,\; a < c < b$

1.7 Removing Discontinuities

1. 1.7.1

   Removing Point Discontinuities

   Redefine the function at the hole so its value equals the limit.

   e.g. $f(x) = \frac{x^2 - 9}{x - 3}$ for $x \ne 3$: limit is $6$, so set $f(3) = 6$

   $\displaystyle f(a) = \lim_{x\to a} f(x)$
2. 1.7.2

   Removing Jump Discontinuities

   Choose the unknown constant so the two pieces meet. Set the left-hand and right-hand limits equal and solve.

   e.g. $f(x) = kx + 1$ for $x < 2$, $x^2$ for $x \ge 2$: $2k + 1 = 4 \Rightarrow k = \tfrac32$

   $\displaystyle \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x)$
3. 1.7.3

   Removing Discontinuities From Rational Functions

   Factor and cancel. A canceled factor $(x - a)$ leaves a hole, which you remove by setting $f(a)$ equal to the simplified function at $a$. A factor left in the denominator gives a vertical asymptote, which cannot be removed.

   e.g. $\frac{x^2 - x - 2}{x^2 - 4} = \frac{x + 1}{x + 2}$ for $x \ne 2$: set $f(2) = \tfrac34$. $x = -2$ stays an asymptote

2

Unit 2 19 topics

Differentiation: Definition and Fundamental Properties

Introduction to Differentiation, and Derivatives of Functions and the Rules of Differentiation.

2.8 Introduction to Differentiation

1. 2.8.1

   The Average Rate of Change of a Function over a Varying Interval

   The average rate of change from $x$ to $x + h$ is the slope of the secant line. The result is an expression in $x$ and $h$.

   e.g. $f(x) = x^2$: $\frac{(x + h)^2 - x^2}{h} = 2x + h$

   $\displaystyle \frac{f(x + h) - f(x)}{h}$
2. 2.8.2

   The Instantaneous Rate of Change of a Function at a Point

   Shrink the interval to zero. The instantaneous rate of change at $a$ is the limit of the average rate, which is the slope of the tangent line.

   e.g. $f(x) = x^2$ at $a = 3$: $\lim_{h\to0}(6 + h) = 6$

   $\displaystyle \lim_{h\to0} \frac{f(a + h) - f(a)}{h}$
3. 2.8.3

   Defining the Derivative Using Derivative Notation

   The derivative $f'(x)$ is the instantaneous rate of change as a function of $x$. Other notations: $y'$, $\frac{dy}{dx}$, $\frac{d}{dx}f(x)$.

   e.g. $f(x) = x^2 + 1$: $f'(x) = \lim_{h\to0}(2x + h) = 2x$

   $\displaystyle f'(x) = \lim_{h\to0} \frac{f(x + h) - f(x)}{h}$

   $\displaystyle f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$
4. 2.8.4

   Continuity and Differentiability of Functions

   If $f$ is differentiable at $a$, it is continuous at $a$. The reverse fails: corners, cusps, and vertical tangents are continuous but not differentiable. A discontinuity always means no derivative.

   e.g. $\lvert x \rvert$ is continuous at $0$ but has a corner there, so $f'(0)$ does not exist

   $\displaystyle \text{differentiable} \Rightarrow \text{continuous}$

   $\displaystyle \text{continuous} \not\Rightarrow \text{differentiable}$
5. 2.8.5

   The Power Rule for Differentiation

   Bring the exponent down and subtract one from it. Works for any real exponent. The derivative of a constant is $0$.

   e.g. $\frac{d}{dx}x^{-2} = -2x^{-3}$, $\frac{d}{dx}\sqrt{x} = \tfrac12 x^{-1/2}$

   $\displaystyle \frac{d}{dx} x^n = n x^{n-1}$

   $\displaystyle \frac{d}{dx} c = 0$
6. 2.8.6

   The Sum and Constant Multiple Rules for Differentiation

   Differentiate term by term. Constant factors stay in front.

   e.g. $\frac{d}{dx}(4x^3 - 5x + 7) = 12x^2 - 5$

   $\displaystyle (f + g)' = f' + g'$

   $\displaystyle (cf)' = c f'$
7. 2.8.7

   Calculating the Slope of a Tangent Line Using Differentiation

   The slope of the tangent line at $x = a$ is $f'(a)$. Differentiate first, then plug in $a$.

   e.g. $f(x) = x^3 - 2x$: $f'(x) = 3x^2 - 2$, so the slope at $x = 2$ is $f'(2) = 10$

   $\displaystyle m = f'(a)$
8. 2.8.8

   Calculating the Equation of a Tangent Line Using Differentiation

   Get the point $(a, f(a))$ and the slope $f'(a)$, then use point-slope form.

   e.g. $f(x) = x^2$ at $x = 3$: $f(3) = 9$, $f'(3) = 6$, so $y = 6x - 9$

   $\displaystyle y - f(a) = f'(a)(x - a)$
9. 2.8.9

   Calculating the Equation of a Normal Line Using Differentiation

   The normal line is perpendicular to the tangent, so its slope is $-\frac{1}{f'(a)}$. If $f'(a) = 0$, the normal line is vertical: $x = a$.

   e.g. $f(x) = x^2$ at $x = 1$: $f'(1) = 2$, so $y - 1 = -\tfrac12(x - 1)$, or $y = -\tfrac12 x + \tfrac32$

   $\displaystyle y - f(a) = -\frac{1}{f'(a)}(x - a)$

2.9 Derivatives of Functions and the Rules of Differentiation

1. 2.9.1

   Differentiating Exponential Functions

   $e^x$ is its own derivative. For any other base $a > 0$, multiply by $\ln a$.

   e.g. $\frac{d}{dx}(3e^x + 2^x) = 3e^x + 2^x \ln 2$

   $\displaystyle \frac{d}{dx} e^x = e^x$

   $\displaystyle \frac{d}{dx} a^x = a^x \ln a$
2. 2.9.2

   Differentiating Logarithmic Functions

   The derivative of $\ln x$ is $\frac1x$. For any other base, divide by $\ln a$ too. Valid for $x > 0$.

   e.g. $\frac{d}{dx}(5\ln x) = \tfrac5x$, $\frac{d}{dx}\log x = \frac{1}{x \ln 10}$

   $\displaystyle \frac{d}{dx} \ln x = \frac{1}{x}$

   $\displaystyle \frac{d}{dx} \log_a x = \frac{1}{x \ln a}$
3. 2.9.3

   Differentiating Trigonometric Functions

   These rules hold only with $x$ in radians. The derivative of cosine has a minus sign.

   e.g. $\frac{d}{dx}(2\sin x - \cos x) = 2\cos x + \sin x$

   $\displaystyle \frac{d}{dx} \sin x = \cos x$

   $\displaystyle \frac{d}{dx} \cos x = -\sin x$

   $\displaystyle \frac{d}{dx} \tan x = \sec^2 x$
4. 2.9.4

   Second and Higher-Order Derivatives

   The second derivative is the derivative of the first derivative. Keep differentiating for higher orders.

   e.g. $f(x) = x^4$: $f' = 4x^3$, $f'' = 12x^2$, $f''' = 24x$

   $\displaystyle f''(x) = \frac{d}{dx} f'(x) = \frac{d^2y}{dx^2}$

   $\displaystyle f^{(n)}(x) = \frac{d^n y}{dx^n}$
5. 2.9.5

   The Product Rule for Differentiation

   Derivative of the first times the second, plus the first times the derivative of the second.

   e.g. $\frac{d}{dx}(x^2 \sin x) = 2x\sin x + x^2 \cos x$

   $\displaystyle (fg)' = f'g + fg'$
6. 2.9.6

   The Quotient Rule for Differentiation

   Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared. The order of the subtraction matters.

   e.g. $\frac{d}{dx}\frac{x}{x + 1} = \frac{(x + 1) - x}{(x + 1)^2} = \frac{1}{(x + 1)^2}$

   $\displaystyle \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$
7. 2.9.7

   Differentiating Reciprocal Trigonometric Functions

   The "co" functions get a minus sign. These rules hold only with $x$ in radians.

   e.g. $\frac{d}{dx}(x\sec x) = \sec x + x\sec x \tan x$

   $\displaystyle \frac{d}{dx} \sec x = \sec x \tan x$

   $\displaystyle \frac{d}{dx} \csc x = -\csc x \cot x$

   $\displaystyle \frac{d}{dx} \cot x = -\csc^2 x$
8. 2.9.8

   Calculating Derivatives From Data and Tables

   Write the rule in symbols, then look up each value of $f$, $g$, $f'$, $g'$ at the given input. With only data points, estimate $f'(a)$ by the slope between nearby points.

   e.g. $h = fg$ with $f(2) = 3$, $f'(2) = 1$, $g(2) = 4$, $g'(2) = -2$: $h'(2) = 1 \cdot 4 + 3(-2) = -2$

9. 2.9.9

   Calculating Derivatives From Graphs

   $f'(a)$ is the slope of the graph at $x = a$. On a straight segment, use the slope of that segment. At a corner or break, $f'(a)$ does not exist.

   e.g. $f$ is a line through $(0, 1)$ and $(2, 5)$: $f'(1) = 2$, so $h = 3f$ has $h'(1) = 6$

10. 2.9.10

    Recognizing Derivatives in Limits

    A limit in difference-quotient form is a derivative. Identify $f$ and $a$, then differentiate instead of evaluating the limit.

    e.g. $\lim_{h\to0}\frac{(2 + h)^3 - 8}{h}$ is $f'(2)$ for $f(x) = x^3$, so it equals $3 \cdot 2^2 = 12$

    $\displaystyle \lim_{h\to0} \frac{f(a + h) - f(a)}{h} = f'(a)$

    $\displaystyle \lim_{x\to a} \frac{f(x) - f(a)}{x - a} = f'(a)$

3

Unit 3 13 topics

Differentiation: Composite, Implicit, and Inverse Functions

Differentiating Composite Functions, and Differentiating Implicit and Inverse Functions.

3.10 Differentiating Composite Functions

1. 3.10.1

   The Chain Rule for Differentiation

   For a composite function, differentiate the outside function (leaving the inside alone), then multiply by the derivative of the inside.

   e.g. $\frac{d}{dx}(3x + 1)^4 = 4(3x + 1)^3 \cdot 3 = 12(3x + 1)^3$

   $\displaystyle \frac{d}{dx} f(g(x)) = f'(g(x))\, g'(x)$

   $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
2. 3.10.2

   The Chain Rule With Exponential Functions

   Keep the exponential as it is, then multiply by the derivative of the exponent.

   e.g. $\frac{d}{dx}e^{x^2} = 2x e^{x^2}$

   $\displaystyle \frac{d}{dx} e^{u} = e^{u}\, u'$

   $\displaystyle \frac{d}{dx} a^{u} = a^{u} \ln a \cdot u'$
3. 3.10.3

   The Chain Rule With Logarithmic Functions

   The derivative of $\ln u$ is the derivative of the inside over the inside.

   e.g. $\frac{d}{dx}\ln(x^2 + 1) = \frac{2x}{x^2 + 1}$

   $\displaystyle \frac{d}{dx} \ln u = \frac{u'}{u}$

   $\displaystyle \frac{d}{dx} \log_a u = \frac{u'}{u \ln a}$
4. 3.10.4

   The Chain Rule With Trigonometric Functions

   Apply the trig rule to the outside, keep the same angle, and multiply by the derivative of the angle. For a power like $\sin^2 x = (\sin x)^2$, the power is the outside.

   e.g. $\frac{d}{dx}\cos 5x = -5\sin 5x$, $\frac{d}{dx}\sin^2 x = 2\sin x \cos x$

   $\displaystyle \frac{d}{dx} \sin u = u' \cos u$

   $\displaystyle \frac{d}{dx} \cos u = -u' \sin u$

   $\displaystyle \frac{d}{dx} \tan u = u' \sec^2 u$
5. 3.10.5

   Calculating Derivatives From Data Using the Chain Rule

   For $h(x) = f(g(x))$, find $g(a)$ first, then look up $f'$ at that value, not at $a$.

   e.g. $g(1) = 3$, $g'(1) = 2$, $f'(3) = 5$: $h'(1) = f'(3) \cdot g'(1) = 10$

   $\displaystyle h'(a) = f'(g(a)) \cdot g'(a)$
6. 3.10.6

   Calculating Derivatives From Graphs Using the Chain Rule

   Read $g(a)$ and the slope $g'(a)$ from the graph of $g$. Then read the slope of $f$ at $x = g(a)$ from the graph of $f$.

   e.g. $g$ is a line through $(0, 0)$ and $(1, 2)$, and $f$ has slope $3$ at $x = 2$: $h'(1) = 3 \cdot 2 = 6$

   $\displaystyle h'(a) = f'(g(a)) \cdot g'(a)$
7. 3.10.7

   Selecting Procedures for Calculating Derivatives

   Find the last operation: a sum uses the sum rule, a product the product rule, a quotient the quotient rule, a composition the chain rule. Rewrite roots and constant-over-power fractions as powers first when you can.

   e.g. $\frac{d}{dx}\frac{3}{x^2} = \frac{d}{dx}3x^{-2} = -6x^{-3}$, no quotient rule needed

3.11 Differentiating Implicit and Inverse Functions

1. 3.11.1

   Implicit Differentiation

   Differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every derivative of a $y$-term picks up a factor of $\frac{dy}{dx}$. Then solve for $\frac{dy}{dx}$.

   e.g. $x^2 + y^2 = 25$: $2x + 2y\,y' = 0 \Rightarrow y' = -\tfrac{x}{y}$, so at $(3, 4)$, $y' = -\tfrac34$

   $\displaystyle \frac{d}{dx}\big[y^n\big] = n y^{n-1}\frac{dy}{dx}$
2. 3.11.2

   Calculating dy/dx Using dx/dy

   When $x$ is given as a function of $y$, find $\frac{dx}{dy}$ and take its reciprocal. This works wherever $\frac{dx}{dy} \ne 0$.

   e.g. $x = y^3 + y$: $\tfrac{dx}{dy} = 3y^2 + 1$, so $\tfrac{dy}{dx} = \tfrac{1}{3y^2 + 1}$, which is $\tfrac14$ at $y = 1$

   $\displaystyle \frac{dy}{dx} = \frac{1}{dx/dy}$
3. 3.11.3

   Differentiating Inverse Functions

   The derivative of $f^{-1}$ at $x$ is the reciprocal of $f'$ evaluated at $f^{-1}(x)$. It requires $f'\big(f^{-1}(x)\big) \ne 0$.

   e.g. $f(x) = e^x$, $f^{-1}(x) = \ln x$: $(f^{-1})'(x) = \tfrac{1}{e^{\ln x}} = \tfrac1x$

   $\displaystyle (f^{-1})'(x) = \frac{1}{f'\big(f^{-1}(x)\big)}$
4. 3.11.4

   Differentiating an Inverse Function at a Point

   To find $(f^{-1})'(b)$, first find the $a$ with $f(a) = b$, then take $\frac{1}{f'(a)}$. Slopes of $f$ and $f^{-1}$ at mirror-image points are reciprocals.

   e.g. $f(x) = x^3 + x$: $f(1) = 2$ and $f'(1) = 4$, so $(f^{-1})'(2) = \tfrac14$

   $\displaystyle f(a) = b$

   $\displaystyle \Rightarrow\; (f^{-1})'(b) = \frac{1}{f'(a)}$
5. 3.11.5

   Differentiating Inverse Trigonometric Functions

   Memorize these three. With an inside function, multiply by its derivative (chain rule). The arcsine and arccosine formulas hold for $-1 < x < 1$.

   e.g. $\tfrac{d}{dx}\tan^{-1}(2x) = \tfrac{2}{1 + 4x^2}$

   $\displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$

   $\displaystyle \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}$

   $\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$
6. 3.11.6

   Differentiating Inverse Reciprocal Trigonometric Functions

   Each "co" function is the negative of its partner. The arcsecant and arccosecant formulas hold for $\lvert x \rvert > 1$.

   e.g. $\tfrac{d}{dx}\cot^{-1}(3x) = -\tfrac{3}{1 + 9x^2}$

   $\displaystyle \frac{d}{dx}\sec^{-1}x = \frac{1}{\lvert x \rvert\sqrt{x^2 - 1}}$

   $\displaystyle \frac{d}{dx}\csc^{-1}x = -\frac{1}{\lvert x \rvert\sqrt{x^2 - 1}}$

   $\displaystyle \frac{d}{dx}\cot^{-1}x = -\frac{1}{1 + x^2}$

4

Unit 4 17 topics

Contextual Applications of Differentiation

Contextual Applications of Differentiation, Estimating Derivatives, Displacement, Velocity, and Acceleration, and Related Rates of Change.

4.12 Contextual Applications of Differentiation

1. 4.12.1

   Interpreting the Meaning of the Derivative in Context

   $f'(a)$ is the instantaneous rate of change of $f$ when the input is $a$. Its units are output units per input unit. A positive value means the quantity is increasing; a negative value means it is decreasing.

   e.g. $V(t)$ = gallons in a tank after $t$ minutes. $V'(5) = -3$ means that at 5 minutes the water is draining at 3 gallons per minute.

   $\displaystyle \text{units of } f' = \frac{\text{units of } f}{\text{units of } t}$
2. 4.12.2

   Rates of Change in Applied Contexts

   For the rate at an instant, differentiate the model and plug in the input. For the rate over an interval, use the average rate of change.

   e.g. $P(t) = 200 + 15t^2$ bacteria: $P'(t) = 30t$, so at $t = 4$ h the colony grows by 120 bacteria per hour

   $\displaystyle \text{average} = \frac{f(b) - f(a)}{b - a}$

   $\displaystyle \text{instantaneous} = f'(a)$

4.13 Estimating Derivatives

1. 4.13.1

   Estimating Derivatives Using a Forward Difference Quotient

   Estimate the slope at $a$ using $a$ and a point a small step $h$ to the right. In a table, use the next data point.

   e.g. $f(2) = 5$, $f(2.5) = 6.5$: $f'(2) \approx \tfrac{6.5 - 5}{0.5} = 3$

   $\displaystyle f'(a) \approx \frac{f(a + h) - f(a)}{h}$
2. 4.13.2

   Estimating Derivatives Using a Backward Difference Quotient

   Estimate the slope at $a$ using $a$ and a point a small step $h$ to the left. In a table, use the previous data point.

   e.g. $f(1.5) = 4$, $f(2) = 5$: $f'(2) \approx \tfrac{5 - 4}{0.5} = 2$

   $\displaystyle f'(a) \approx \frac{f(a) - f(a - h)}{h}$
3. 4.13.3

   Estimating Derivatives Using a Central Difference Quotient

   Use the points a step $h$ on each side of $a$. The gap in inputs is $2h$. This is usually the most accurate of the three estimates.

   e.g. $f(1.5) = 4$, $f(2.5) = 6.5$: $f'(2) \approx \tfrac{6.5 - 4}{1} = 2.5$

   $\displaystyle f'(a) \approx \frac{f(a + h) - f(a - h)}{2h}$

4.14 Displacement, Velocity, and Acceleration

1. 4.14.1

   Calculating Velocity for Straight-Line Motion Using Differentiation

   Velocity is the derivative of position. Speed is the absolute value of velocity.

   e.g. $s(t) = t^3 - 6t$: $v(t) = 3t^2 - 6$, so $v(2) = 6$ units per second

   $\displaystyle v(t) = s'(t)$

   $\displaystyle \text{speed} = \lvert v(t) \rvert$
2. 4.14.2

   Determining Characteristics of Moving Objects Using Differentiation

   The object is at rest when $v = 0$, moves right (or up) when $v > 0$, moves left (or down) when $v < 0$, and changes direction where $v$ changes sign. It speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs.

   e.g. $s(t) = t^2 - 4t$: $v = 2t - 4$, $a = 2$. At rest at $t = 2$. At $t = 1$, $v = -2$ and $a = 2$, so it moves left and slows down.

   $\displaystyle v = 0 \;\to\; \text{at rest}$

   $\displaystyle v \cdot a > 0 \;\to\; \text{speeding up}$

   $\displaystyle v \cdot a < 0 \;\to\; \text{slowing down}$
3. 4.14.3

   Calculating Acceleration for Straight-Line Motion Using Differentiation

   Acceleration is the derivative of velocity, which is the second derivative of position. Units are distance per time squared.

   e.g. $s(t) = t^3 - 6t$: $v(t) = 3t^2 - 6$, $a(t) = 6t$, so $a(2) = 12$

   $\displaystyle a(t) = v'(t) = s''(t)$

4.15 Related Rates of Change

1. 4.15.1

   Introduction to Related Rates

   Write an equation linking the quantities, differentiate both sides with respect to $t$, then substitute the known values and solve for the unknown rate. Substitute only after differentiating.

   e.g. $y = x^2$, $\tfrac{dx}{dt} = 3$ when $x = 2$: $\tfrac{dy}{dt} = 2x\tfrac{dx}{dt} = 2(2)(3) = 12$

   $\displaystyle \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
2. 4.15.2

   Related Rates With Implicit Functions

   When the equation is not solved for one variable, differentiate every term with respect to $t$. Each $x$-term gets a $\frac{dx}{dt}$ and each $y$-term gets a $\frac{dy}{dt}$.

   e.g. $x^2 + y^2 = 25$ at $(3, 4)$ with $\tfrac{dx}{dt} = 2$: $2(3)(2) + 2(4)\tfrac{dy}{dt} = 0 \Rightarrow \tfrac{dy}{dt} = -\tfrac32$

   $\displaystyle \frac{d}{dt}\big[y^2\big] = 2y\frac{dy}{dt}$
3. 4.15.3

   Calculating Related Rates With Circles and Spheres

   Differentiate $A = \pi r^2$, $C = 2\pi r$, or $V = \frac43\pi r^3$ with respect to $t$.

   e.g. Balloon radius grows 2 cm/s. At $r = 5$: $\tfrac{dV}{dt} = 4\pi(25)(2) = 200\pi$ cm³/s

   $\displaystyle \frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

   $\displaystyle \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$
4. 4.15.4

   Calculating Related Rates With Squares

   Differentiate the area $A = s^2$ or perimeter $P = 4s$ with respect to $t$.

   e.g. Side grows 3 in/min. At $s = 4$: $\tfrac{dA}{dt} = 2(4)(3) = 24$ in²/min

   $\displaystyle \frac{dA}{dt} = 2s\frac{ds}{dt}$

   $\displaystyle \frac{dP}{dt} = 4\frac{ds}{dt}$
5. 4.15.5

   Calculating Related Rates With Rectangular Solids

   Differentiate $V = lwh$ with the product rule, so each term has one changing dimension. Primes below mean $\frac{d}{dt}$. For a cube, $V = s^3$.

   e.g. $l = 2$, $w = 3$, $h = 4$ with $l' = 1$, $w' = 0$, $h' = -1$: $V' = 12 + 0 - 6 = 6$

   $\displaystyle V' = l'wh + lw'h + lwh'$

   $\displaystyle \text{cube: } V' = 3s^2 s'$
6. 4.15.6

   Calculating Related Rates Using the Pythagorean Theorem

   For a right triangle with legs $x$, $y$ and hypotenuse $z$, differentiate $x^2 + y^2 = z^2$ with respect to $t$. A fixed length has rate 0.

   e.g. 10 ft ladder, base 6 ft out sliding away at 2 ft/s: top is 8 ft up, $6(2) + 8\tfrac{dy}{dt} = 0 \Rightarrow \tfrac{dy}{dt} = -\tfrac32$ ft/s

   $\displaystyle x^2 + y^2 = z^2$

   $\displaystyle \Rightarrow\; x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$
7. 4.15.7

   Calculating Related Rates Using Similar Triangles

   Use a proportion between similar triangles to write one length in terms of another, then differentiate. Common in shadow and water-tank problems.

   e.g. 6 ft person walks from a 15 ft lamppost at 5 ft/s. $\tfrac{15}{x + s} = \tfrac{6}{s} \Rightarrow s = \tfrac23 x$, so $\tfrac{ds}{dt} = \tfrac23(5) = \tfrac{10}{3}$ ft/s

   $\displaystyle \frac{\text{height}_1}{\text{base}_1} = \frac{\text{height}_2}{\text{base}_2}$
8. 4.15.8

   Calculating Related Rates Using Trigonometry

   Relate the angle to the sides with sin, cos, or tan, then differentiate with respect to $t$ using the chain rule. Angles must be in radians.

   e.g. Balloon 100 m away rises 5 m/s. $\tan\theta = \tfrac{h}{100}$. At $h = 100$, $\theta = \tfrac{\pi}{4}$: $2\tfrac{d\theta}{dt} = \tfrac{5}{100} \Rightarrow \tfrac{d\theta}{dt} = \tfrac{1}{40}$ rad/s

   $\displaystyle \frac{d}{dt}\sin\theta = \cos\theta\,\frac{d\theta}{dt}$

   $\displaystyle \frac{d}{dt}\tan\theta = \sec^2\theta\,\frac{d\theta}{dt}$
9. 4.15.9

   Calculating Related Rates With Cones

   Use similar triangles to write $r$ in terms of $h$ (the ratio $\frac{r}{h}$ matches the full cone's), substitute into the volume formula, then differentiate.

   e.g. Tank with $R = 3$, $H = 6$: $r = \tfrac{h}{2}$, $V = \tfrac{\pi h^3}{12}$, $\tfrac{dV}{dt} = \tfrac{\pi h^2}{4}\tfrac{dh}{dt}$. Filling at 2 ft³/min, at $h = 4$: $\tfrac{dh}{dt} = \tfrac{1}{2\pi}$ ft/min

   $\displaystyle V = \tfrac13\pi r^2 h$

   $\displaystyle \frac{r}{h} = \frac{R}{H}$

5

Unit 5 29 topics

Analytical Applications of Differentiation

L'Hopital's Rule, Analytical Applications of Differentiation, Analysis of Curves, and more.

5.16 L'Hopital's Rule

1. 5.16.1

   L'Hopital's Rule

   If direct substitution gives $\frac00$ or $\frac{\infty}{\infty}$, the limit of $\frac{f}{g}$ equals the limit of $\frac{f'}{g'}$, provided that limit exists (or is $\pm\infty$). Differentiate top and bottom separately, not with the quotient rule, and only after confirming the form.

   e.g. $\lim_{x \to 0}\tfrac{\sin x}{x}$ is $\tfrac00$, so it equals $\lim_{x \to 0}\tfrac{\cos x}{1} = 1$

   $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$
2. 5.16.2

   L'Hopital's Rule Applied to Tables

   Check from the table that $f(a)$ and $g(a)$ are both 0. If so, and $f'$, $g'$ are continuous with $g'(a) \ne 0$, the limit is the ratio of the derivative values.

   e.g. $f(2) = g(2) = 0$, $f'(2) = 6$, $g'(2) = 4$: $\lim_{x \to 2}\tfrac{f(x)}{g(x)} = \tfrac64 = \tfrac32$

   $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}$

5.17 Analytical Applications of Differentiation

1. 5.17.1

   The Mean Value Theorem

   If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then at some $c$ in $(a, b)$ the instantaneous slope equals the average slope.

   e.g. $f(x) = x^2$ on $[1, 3]$: average slope $\tfrac{9 - 1}{2} = 4$, and $f'(c) = 2c = 4 \Rightarrow c = 2$

   $\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$
2. 5.17.2

   Global vs. Local Extrema and Critical Points

   A critical point is an $x$ in the domain where $f'(x) = 0$ or $f'(x)$ is undefined. Interior local extrema occur only at critical points, but not every critical point is one. Global (absolute) extrema are the highest and lowest values overall and can also occur at endpoints.

   e.g. $f(x) = x^3$: $f'(0) = 0$, but $x = 0$ is not an extremum

   $\displaystyle f'(c) = 0 \;\text{or}\; f'(c) \;\text{undefined}$
3. 5.17.3

   The Extreme Value Theorem

   If $f$ is continuous on a closed interval $[a, b]$, then $f$ has both an absolute maximum and an absolute minimum on $[a, b]$. Both conditions are required.

   e.g. $f(x) = \tfrac1x$ on $(0, 1]$ has no maximum because the interval is not closed

4. 5.17.4

   Using Differentiation to Calculate Critical Points

   Find $f'(x)$. Solve $f'(x) = 0$ and find where $f'(x)$ is undefined, keeping only $x$-values in the domain of $f$.

   e.g. $f(x) = x^3 - 3x^2$: $f'(x) = 3x(x - 2)$, critical points $x = 0$ and $x = 2$

   $\displaystyle f'(x) = 0 \;\text{or undefined}$
5. 5.17.5

   Determining Intervals on Which a Function Is Increasing or Decreasing

   Split the number line at the critical points and test the sign of $f'$ in each interval. $f$ increases where $f' > 0$ and decreases where $f' < 0$.

   e.g. $f'(x) = 3x(x - 2)$: increasing on $(-\infty, 0)$ and $(2, \infty)$, decreasing on $(0, 2)$

   $\displaystyle f' > 0 \;\Rightarrow\; \text{increasing}$

   $\displaystyle f' < 0 \;\Rightarrow\; \text{decreasing}$
6. 5.17.6

   Using the First Derivative Test to Classify Local Extrema

   At a critical point $c$ where $f$ is continuous, check the sign of $f'$ on each side. No sign change means no extremum.

   e.g. $f(x) = x^3 - 3x^2$: local max $f(0) = 0$, local min $f(2) = -4$

   $\displaystyle f'\!: + \to - \;\Rightarrow\; \text{local max}$

   $\displaystyle f'\!: - \to + \;\Rightarrow\; \text{local min}$
7. 5.17.7

   The Candidates Test

   For the absolute extrema of a continuous $f$ on $[a, b]$, evaluate $f$ at the critical points inside the interval and at both endpoints. The largest value is the absolute max; the smallest is the absolute min.

   e.g. $f(x) = x^3 - 3x^2$ on $[1, 4]$: $f(1) = -2$, $f(2) = -4$, $f(4) = 16$. Max 16, min $-4$

8. 5.17.8

   Intervals of Concavity

   $f$ is concave up where $f'$ is increasing (curve bends like a cup) and concave down where $f'$ is decreasing (curve bends like a cap).

   e.g. $f(x) = x^3$: slopes $3x^2$ fall on $(-\infty, 0)$ and rise on $(0, \infty)$, so concave down, then concave up

   $\displaystyle f' \text{ increasing} \;\Rightarrow\; \text{concave up}$

   $\displaystyle f' \text{ decreasing} \;\Rightarrow\; \text{concave down}$
9. 5.17.9

   Relating Concavity to the Second Derivative

   The sign of $f''$ gives the concavity. Find where $f'' = 0$ or is undefined, then test the sign in each interval.

   e.g. $f(x) = x^3 - 3x^2$: $f''(x) = 6x - 6$, concave down on $(-\infty, 1)$, concave up on $(1, \infty)$

   $\displaystyle f'' > 0 \;\Rightarrow\; \text{concave up}$

   $\displaystyle f'' < 0 \;\Rightarrow\; \text{concave down}$
10. 5.17.10

    Points of Inflection

    A point on the graph where $f$ is continuous and the concavity changes, so $f''$ changes sign. $f''(c) = 0$ alone is not enough: $x^4$ has $f''(0) = 0$ but no inflection point.

    e.g. $f(x) = x^3 - 3x^2$: $f'' = 6x - 6$ changes sign at $x = 1$, so $(1, -2)$ is an inflection point

11. 5.17.11

    The Second Derivative Test

    At a critical point $c$ with $f'(c) = 0$, the sign of $f''(c)$ classifies it. If $f''(c) = 0$, the test is inconclusive; use the first derivative test.

    e.g. $f(x) = x^3 - 3x^2$: $f''(0) = -6$ (local max), $f''(2) = 6$ (local min)

    $\displaystyle f''(c) > 0 \;\Rightarrow\; \text{local min}$

    $\displaystyle f''(c) < 0 \;\Rightarrow\; \text{local max}$

5.18 Analysis of Curves

1. 5.18.1

   Sketching the Derivative of a Function From the Function's Graph

   The height of $f'$ at $x$ is the slope of $f$ at $x$. Horizontal tangents on $f$ become zeros of $f'$, and $f'$ is above the axis where $f$ rises and below where $f$ falls. At corners and cusps, $f'$ is undefined.

   e.g. $f$ is an upward parabola with vertex at $x = 1$: $f'$ is a rising line through $(1, 0)$

2. 5.18.2

   Interpreting the Graph of a Function's Derivative

   Where the graph of $f'$ is above the $x$-axis, $f$ is increasing; below, $f$ is decreasing. If $f'$ crosses zero going down, $f$ has a local max; going up, a local min; touching zero without crossing gives no extremum.

   e.g. $f'$ crosses the $x$-axis from below to above at $x = 3$, so $f$ has a local min at $x = 3$

3. 5.18.3

   Interpreting the Graph of a Function's Derivative: Concavity and Points of Inflection

   Where the graph of $f'$ rises, $f$ is concave up; where it falls, $f$ is concave down. Peaks and valleys of $f'$ are inflection points of $f$.

   e.g. $f'$ has a peak at $x = 2$, so $f$ switches from concave up to concave down there: an inflection point

4. 5.18.4

   Sketching a Function From the Graph of its Derivative

   Read the sign of $f'$ for where $f$ rises or falls, and whether $f'$ rises or falls for concavity. A given point fixes the height; without one, any vertical shift works.

   e.g. $f'(x) = 2x$ with $f(0) = 1$: $f$ falls, then rises, with a min at $x = 0$ and always concave up, so $f(x) = x^2 + 1$

5. 5.18.5

   Sketching a Function Given Some Derivative Properties

   Translate each condition: $f(a) = b$ is a point, $f'(a) = 0$ is a horizontal tangent, the sign of $f'$ gives rising or falling, and the sign of $f''$ gives concavity. Build the sketch interval by interval.

   e.g. $f(0) = 2$, $f'(0) = 0$, $f''(x) < 0$ everywhere: a hill with its peak at $(0, 2)$

5.19 Approximating Values of a Function

1. 5.19.1

   Approximating Functions Using Local Linearity and Linearization

   Near $x = a$ the tangent line stays close to the curve, so use it to estimate $f(x)$. The estimate is too low where $f$ is concave up and too high where $f$ is concave down.

   e.g. $f(x) = x^3$ at $a = 2$: $L(x) = 8 + 12(x - 2)$, so $2.1^3 \approx 8 + 12(0.1) = 9.2$

   $\displaystyle L(x) = f(a) + f'(a)(x - a)$

   $\displaystyle f(x) \approx L(x) \text{ for } x \text{ near } a$
2. 5.19.2

   Approximating the Roots of a Number Using Local Linearity

   Pick the nearest perfect square (or cube) as $a$, then linearize the root function there.

   e.g. $\sqrt{26} \approx \sqrt{25} + \tfrac{1}{10}(1) = 5.1$

   $\displaystyle \sqrt{x} \approx \sqrt{a} + \frac{1}{2\sqrt{a}}(x - a)$

   $\displaystyle \sqrt[3]{x} \approx \sqrt[3]{a} + \frac{x - a}{3\sqrt[3]{a^2}}$
3. 5.19.3

   Approximating Trigonometric Functions Using Local Linearity

   Linearize at a nearby angle whose values you know. Work in radians, so convert any change in degrees using $1^\circ = \tfrac{\pi}{180}$.

   e.g. $\sin 31^\circ \approx \sin 30^\circ + \cos 30^\circ \cdot \tfrac{\pi}{180} = \tfrac12 + \tfrac{\sqrt3}{2}\cdot\tfrac{\pi}{180} \approx 0.515$

   $\displaystyle \sin x \approx \sin a + \cos a\,(x - a)$

   $\displaystyle \text{near } 0:\; \sin x \approx x,\; \tan x \approx x$

   $\displaystyle \text{near } 0:\; \cos x \approx 1$

5.20 Optimization

1. 5.20.1

   Solving Optimization Problems Using Derivatives

   Write the quantity to optimize as a function of one variable, using the constraint. Find where its derivative is zero or undefined, then check those points and the endpoints of the domain.

   e.g. Two positive numbers sum to 20. $P = x(20 - x)$, $P' = 20 - 2x = 0 \Rightarrow x = 10$, max product $100$

   $\displaystyle f'(x) = 0 \text{ or undefined}$

   $\displaystyle f''(c) < 0 \Rightarrow \text{max at } c$

   $\displaystyle f''(c) > 0 \Rightarrow \text{min at } c$
2. 5.20.2

   Optimization Problems Involving Sectors of Circles

   Use the sector formulas with $\theta$ in radians. Solve the constraint (often a fixed perimeter) for $\theta$, substitute, and optimize in terms of $r$.

   e.g. Perimeter 20: $\theta = \tfrac{20 - 2r}{r}$, $A = 10r - r^2$, $A' = 0 \Rightarrow r = 5$, $\theta = 2$, max area $25$

   $\displaystyle A = \tfrac12 r^2 \theta$

   $\displaystyle s = r\theta$

   $\displaystyle P = 2r + r\theta$
3. 5.20.3

   Optimization Problems Involving Boxes and Trays

   Cut a square of side $x$ from each corner of an $L \times W$ sheet and fold up the sides. The domain is $0 < x < \tfrac{W}{2}$ for the shorter side $W$.

   e.g. $12 \times 12$ sheet: $V = x(12 - 2x)^2$, $V' = (12 - 2x)(12 - 6x) = 0 \Rightarrow x = 2$, max $V = 128$

   $\displaystyle V = x(L - 2x)(W - 2x)$
4. 5.20.4

   Optimization Problems Involving Cylinders

   Use the fixed quantity to write $h$ in terms of $r$, then optimize. For a closed can with fixed volume, the least surface area occurs when $h = 2r$.

   e.g. $V = 16\pi$: $h = \tfrac{16}{r^2}$, $S = 2\pi r^2 + \tfrac{32\pi}{r}$, $S' = 0 \Rightarrow r = 2$, $h = 4$, $S = 24\pi$

   $\displaystyle V = \pi r^2 h$

   $\displaystyle S = 2\pi r^2 + 2\pi r h \;\text{(closed)}$
5. 5.20.5

   Optimizing Distances

   Write the distance with the distance formula in terms of one variable (often time $t$). Minimize $D^2$ instead of $D$: it has the same minimum point and no square root.

   e.g. Car A at $(6 - t, 0)$, car B at $(0, t)$: $D^2 = (6 - t)^2 + t^2$, derivative $4t - 12 = 0 \Rightarrow t = 3$, $D = 3\sqrt2$

   $\displaystyle D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
6. 5.20.6

   Optimizing Distances to Curves

   A point on $y = f(x)$ is $(x, f(x))$. Minimize the squared distance from it to the fixed point $(a, b)$.

   e.g. Closest point on $y = \sqrt{x}$ to $(3, 0)$: $D^2 = (x - 3)^2 + x$, $2x - 5 = 0 \Rightarrow x = \tfrac52$, $D = \tfrac{\sqrt{11}}{2}$

   $\displaystyle D^2 = (x - a)^2 + (f(x) - b)^2$
7. 5.20.7

   Optimization Problems With Inscribed Shapes

   Put a corner of the inscribed shape on the curve at $(x, f(x))$ and write the area or volume in terms of $x$. Use symmetry to double a half.

   e.g. Rectangle under $y = 12 - x^2$: $A = 24x - 2x^3$, $A' = 24 - 6x^2 = 0 \Rightarrow x = 2$, max $A = 32$

   $\displaystyle A = 2x\,f(x) \;\text{(rectangle, symmetric)}$
8. 5.20.8

   Optimization Problems in Economics

   Profit is revenue minus cost. Maximum profit occurs where $P'(x) = 0$, which is where marginal revenue equals marginal cost.

   e.g. $R = 50x - 0.5x^2$, $C = 10x + 100$: $P = 40x - 0.5x^2 - 100$, $P' = 40 - x = 0 \Rightarrow x = 40$, profit $700

   $\displaystyle P(x) = R(x) - C(x)$

   $\displaystyle R(x) = p \cdot x$

   $\displaystyle \text{max profit} \Rightarrow R'(x) = C'(x)$

6

Unit 6 48 topics

Integration

Indefinite Integrals, Approximating Areas with Riemann Sums, Definite Integrals, and more.

6.21 Indefinite Integrals

1. 6.21.1

   The Antiderivative

   $F$ is an antiderivative of $f$ if $F' = f$. Every antiderivative differs by a constant, so add $+\,C$. The power rule works for every $n$ except $-1$.

   e.g. $\int x^3\,dx = \tfrac{x^4}{4} + C$

   $\displaystyle \int f(x)\,dx = F(x) + C$

   $\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1} + C$

   $\displaystyle n \ne -1$
2. 6.21.2

   The Constant Multiple Rule for Indefinite Integrals

   A constant factor can be pulled out in front of the integral.

   e.g. $\int 6x^2\,dx = 6 \cdot \tfrac{x^3}{3} + C = 2x^3 + C$

   $\displaystyle \int k\,f(x)\,dx = k \int f(x)\,dx$
3. 6.21.3

   The Sum Rule for Indefinite Integrals

   Integrate term by term. One $+\,C$ at the end covers all the terms.

   e.g. $\int (4x^3 + 2x - 5)\,dx = x^4 + x^2 - 5x + C$

   $\displaystyle \int (f \pm g)\,dx = \int f\,dx \pm \int g\,dx$
4. 6.21.4

   Integrating the Reciprocal Function

   This is the $n = -1$ case the power rule misses. The absolute value lets it work for negative $x$.

   e.g. $\int \tfrac{3}{x}\,dx = 3 \ln \lvert x \rvert + C$

   $\displaystyle \int \frac{1}{x}\,dx = \ln \lvert x \rvert + C$
5. 6.21.5

   Integrating Exponential Functions

   $e^x$ is its own antiderivative. For any other base $a > 0$ with $a \ne 1$, divide by $\ln a$.

   e.g. $\int (e^x + 2^x)\,dx = e^x + \tfrac{2^x}{\ln 2} + C$

   $\displaystyle \int e^x\,dx = e^x + C$

   $\displaystyle \int a^x\,dx = \frac{a^x}{\ln a} + C$
6. 6.21.6

   Integrating Trigonometric Functions

   Reverse the derivative rules. Watch the signs: the integral of $\sin x$ is negative.

   e.g. $\int (3\cos x - \sec^2 x)\,dx = 3\sin x - \tan x + C$

   $\displaystyle \int \cos x\,dx = \sin x + C$

   $\displaystyle \int \sin x\,dx = -\cos x + C$

   $\displaystyle \int \sec^2 x\,dx = \tan x + C$

   $\displaystyle \int \csc^2 x\,dx = -\cot x + C$

   $\displaystyle \int \sec x \tan x\,dx = \sec x + C$

   $\displaystyle \int \csc x \cot x\,dx = -\csc x + C$
7. 6.21.7

   Integration Using Inverse Trigonometric Functions

   Recognize these two forms. With a constant $a^2$ in place of $1$ (take $a > 0$), divide $x$ by $a$, and add a factor $\tfrac1a$ for arctan.

   e.g. $\int \tfrac{dx}{4 + x^2} = \tfrac12 \arctan \tfrac{x}{2} + C$

   $\displaystyle \int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C$

   $\displaystyle \int \frac{dx}{1 + x^2} = \arctan x + C$

   $\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C$

   $\displaystyle \int \frac{dx}{a^2 + x^2} = \frac1a \arctan \frac{x}{a} + C$

6.22 Approximating Areas with Riemann Sums

1. 6.22.1

   Approximating Areas With the Left Riemann Sum

   Split $[a, b]$ into $n$ equal widths and use the left endpoint of each piece for the height. It underestimates for increasing $f$ and overestimates for decreasing $f$.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 2$: $L_2 = 1\,[f(0) + f(1)] = 1$

   $\displaystyle \Delta x = \frac{b - a}{n}$

   $\displaystyle L_n = \Delta x\,[f(x_0) + \cdots + f(x_{n-1})]$
2. 6.22.2

   Approximating Areas With the Right Riemann Sum

   Use the right endpoint of each piece for the height. It overestimates for increasing $f$ and underestimates for decreasing $f$.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 2$: $R_2 = 1\,[f(1) + f(2)] = 5$

   $\displaystyle \Delta x = \frac{b - a}{n}$

   $\displaystyle R_n = \Delta x\,[f(x_1) + \cdots + f(x_n)]$
3. 6.22.3

   Approximating Areas With the Midpoint Riemann Sum

   Use the midpoint of each piece for the height. It underestimates when $f$ is concave up and overestimates when $f$ is concave down.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 2$: $M_2 = 1\,[f(0.5) + f(1.5)] = 0.25 + 2.25 = 2.5$

   $\displaystyle M_n = \Delta x\,[f(m_1) + \cdots + f(m_n)]$

   $\displaystyle m_i = \text{midpoint of piece } i$
4. 6.22.4

   Approximating Areas With the Trapezoidal Rule

   Use a trapezoid on each piece. The interior heights count twice. The result equals the average of the left and right sums. It overestimates when $f$ is concave up and underestimates when $f$ is concave down.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 2$: $T_2 = \tfrac12[0 + 2(1) + 4] = 3$

   $\displaystyle T_n = \frac{\Delta x}{2}\big[f(x_0) + 2f(x_1)$

   $\displaystyle + \cdots + 2f(x_{n-1}) + f(x_n)\big]$

   $\displaystyle T_n = \frac{L_n + R_n}{2}$
5. 6.22.5

   Left and Right Riemann Sums in Sigma Notation

   The $i$th point is $x_i = a + i\,\Delta x$. The left sum runs $i = 0$ to $n - 1$, and the right sum runs $i = 1$ to $n$.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 4$: $R_4 = \sum_{i=1}^{4} \left(\tfrac{i}{2}\right)^2 \tfrac12 = \tfrac{1 + 4 + 9 + 16}{8} = 3.75$

   $\displaystyle L_n = \sum_{i=0}^{n-1} f(a + i\,\Delta x)\,\Delta x$

   $\displaystyle R_n = \sum_{i=1}^{n} f(a + i\,\Delta x)\,\Delta x$
6. 6.22.6

   Midpoint and Trapezoidal Rules in Sigma Notation

   The $i$th midpoint is $a + (i - \tfrac12)\Delta x$. Each trapezoid averages the heights at its two ends.

   e.g. $f(x) = x^2$ on $[0, 2]$, $n = 2$: $M_2 = (0.5)^2 + (1.5)^2 = 2.5$

   $\displaystyle M_n = \sum_{i=1}^{n} f\big(a + (i - \tfrac12)\Delta x\big)\Delta x$

   $\displaystyle T_n = \sum_{i=1}^{n} \frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x$
7. 6.22.7

   Approximating Areas Under Graphs of Composite Functions

   The method does not change. At each sample point, evaluate the inside function first, then the outside function.

   e.g. $f(x) = (2x - 1)^2$ on $[0, 2]$, $n = 2$, midpoints: $M_2 = 1\,[f(0.5) + f(1.5)] = 0 + 4 = 4$

6.23 Definite Integrals

1. 6.23.1

   Defining Definite Integrals Using Left and Right Riemann Sums

   The definite integral is the limit of Riemann sums as the number of pieces goes to infinity. Left and right sums give the same limit for a continuous $f$.

   e.g. $\lim_{n \to \infty} \sum_{i=1}^{n} \left(\tfrac{i}{n}\right)^2 \tfrac1n = \int_0^1 x^2\,dx = \tfrac13$

   $\displaystyle \int_a^b f(x)\,dx$

   $\displaystyle = \lim_{n \to \infty} \sum_{i=1}^{n} f(a + i\,\Delta x)\,\Delta x$

   $\displaystyle \Delta x = \frac{b - a}{n}$
2. 6.23.2

   The Fundamental Theorem of Calculus

   To evaluate a definite integral of a continuous $f$, find any antiderivative $F$ and subtract its values at the limits. The $+\,C$ cancels.

   e.g. $\int_1^3 2x\,dx = 3^2 - 1^2 = 8$

   $\displaystyle \int_a^b f(x)\,dx = F(b) - F(a)$

   $\displaystyle F'(x) = f(x)$
3. 6.23.3

   Applying the Fundamental Theorem of Calculus to Exponential and Trigonometric

   Use the exponential and trig antiderivatives, then subtract. Trig limits are in radians.

   e.g. $\int_0^{\pi} \sin x\,dx = -\cos \pi + \cos 0 = 1 + 1 = 2$

   $\displaystyle \int_a^b e^x\,dx = e^b - e^a$

   $\displaystyle \int_a^b \cos x\,dx = \sin b - \sin a$

   $\displaystyle \int_a^b \sin x\,dx = -\cos b + \cos a$
4. 6.23.4

   The Sum and Constant Multiple Rules for Definite Integrals

   Definite integrals split across sums and differences, and constants factor out.

   e.g. $\int_0^2 f = 3$ and $\int_0^2 g = 5 \Rightarrow \int_0^2 (2f - g) = 6 - 5 = 1$

   $\displaystyle \int_a^b (f \pm g)\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$

   $\displaystyle \int_a^b k\,f\,dx = k \int_a^b f\,dx$
5. 6.23.5

   Properties of Definite Integrals Involving the Limits of Integration

   Equal limits give zero. Swapping the limits flips the sign. Adjacent intervals add.

   e.g. $\int_1^5 f = 7$ and $\int_1^3 f = 2 \Rightarrow \int_3^5 f = 5$ and $\int_5^3 f = -5$

   $\displaystyle \int_a^a f(x)\,dx = 0$

   $\displaystyle \int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$

   $\displaystyle \int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx$

6.24 The Area Under a Curve

1. 6.24.1

   The Area Bounded by a Curve and the X-Axis

   If $f(x) \ge 0$ on $[a, b]$, the area between the curve and the $x$-axis is the definite integral. Use the $x$-intercepts as limits when no interval is given.

   e.g. $y = 4 - x^2$: $\int_{-2}^{2} (4 - x^2)\,dx = \left(8 - \tfrac83\right) - \left(-8 + \tfrac83\right) = \tfrac{32}{3}$

   $\displaystyle A = \int_a^b f(x)\,dx \quad (f \ge 0)$
2. 6.24.2

   Evaluating Definite Integrals Using Symmetry

   On an interval symmetric about $0$, an even function gives twice the right half and an odd function gives zero.

   e.g. $\int_{-2}^{2} (x^3 + 3x^2)\,dx = 0 + 2\int_0^2 3x^2\,dx = 2(8) = 16$

   $\displaystyle f \text{ even}: \int_{-a}^{a} f\,dx = 2\int_0^a f\,dx$

   $\displaystyle f \text{ odd}: \int_{-a}^{a} f\,dx = 0$
3. 6.24.3

   Finding the Area Between a Curve and the X-Axis When They Intersect

   Area counts every region as positive, but the integral counts regions below the axis as negative. Split at the $x$-intercepts and add the absolute values.

   e.g. $y = x^2 - 1$ on $[0, 2]$: $\int_0^1 = -\tfrac23$, $\int_1^2 = \tfrac43$, so area $= \tfrac23 + \tfrac43 = 2$

   $\displaystyle A = \int_a^b \lvert f(x) \rvert\,dx$
4. 6.24.4

   The Area Bounded by a Curve and the Y-Axis

   Solve for $x$ in terms of $y$ and integrate with respect to $y$ between the $y$-limits.

   e.g. Region between $y = \sqrt{x}$, the $y$-axis, and $y = 2$: $x = y^2$, $A = \int_0^2 y^2\,dy = \tfrac83$

   $\displaystyle A = \int_c^d g(y)\,dy \quad (x = g(y) \ge 0)$
5. 6.24.5

   Calculating the Definite Integral of a Function Given Its Graph

   Break the region into triangles, rectangles, and circle parts and use geometry. Area above the $x$-axis counts positive and area below counts negative.

   e.g. $f$ is the line through $(0, 2)$ and $(4, -2)$: $\int_0^3 f\,dx = 2 - \tfrac12 = \tfrac32$

   $\displaystyle \int_a^b f\,dx = A_{\text{above}} - A_{\text{below}}$
6. 6.24.6

   Calculating the Definite Integral of a Function's Derivative Given its Graph

   The integral of $f'$ is the net change in $f$. Read $f(a)$ and $f(b)$ off the graph of $f$ and subtract.

   e.g. Graph of $f$ passes through $(1, 4)$ and $(5, -2)$: $\int_1^5 f'(x)\,dx = -2 - 4 = -6$

   $\displaystyle \int_a^b f'(x)\,dx = f(b) - f(a)$
7. 6.24.7

   Definite Integrals of Piecewise Functions

   Split the integral at each break point and integrate each piece only over its own interval.

   e.g. $f(x) = 3x^2$ for $x < 1$ and $f(x) = 2$ for $x \ge 1$: $\int_0^3 f\,dx = 1 + 4 = 5$

   $\displaystyle \int_a^c f\,dx = \int_a^b f_1\,dx + \int_b^c f_2\,dx$

6.25 Accumulation Functions

1. 6.25.1

   The Integral as an Accumulation Function

   $F(x)$ is the signed area under $f$ from a fixed start $a$ to a variable endpoint $x$. It starts at $F(a) = 0$, grows where $f > 0$, and shrinks where $f < 0$.

   e.g. $F(x) = \int_0^x 2t\,dt = x^2$, so $F(3) = 9$

   $\displaystyle F(x) = \int_a^x f(t)\,dt$
2. 6.25.2

   The Second Fundamental Theorem of Calculus

   The derivative of an accumulation function is the integrand evaluated at the upper limit. If the upper limit is $g(x)$, multiply by $g'(x)$ (chain rule).

   e.g. $\frac{d}{dx}\int_1^{x^2} \cos t\,dt = 2x\cos(x^2)$

   $\displaystyle \frac{d}{dx}\int_a^x f(t)\,dt = f(x)$

   $\displaystyle \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$
3. 6.25.3

   Maximizing a Function Using the Graph of Its Derivative

   $f$ has a relative max where $f'$ crosses the axis from positive to negative. For an absolute max on $[a, b]$, compare those points with the endpoints.

   e.g. $f' > 0$ on $(0, 3)$ and $f' < 0$ on $(3, 5)$: relative max at $x = 3$

   $\displaystyle f' \colon + \to - \;\Rightarrow\; \text{relative max}$
4. 6.25.4

   Minimizing a Function Using the Graph of its Derivative

   $f$ has a relative min where $f'$ crosses the axis from negative to positive. For an absolute min on $[a, b]$, compare those points with the endpoints.

   e.g. $f' < 0$ on $(-2, 1)$ and $f' > 0$ on $(1, 4)$: relative min at $x = 1$

   $\displaystyle f' \colon - \to + \;\Rightarrow\; \text{relative min}$
5. 6.25.5

   Further Optimizing Functions Using Graphs of Derivatives

   Find $f$ at each candidate by adding the signed area under $f'$ to a known value. Then compare the critical points and endpoints to get the absolute max and min.

   e.g. $f(0) = 2$, area $5$ above the axis on $[0, 3]$, area $7$ below on $[3, 6]$: $f(3) = 7$ (abs max), $f(6) = 0$ (abs min)

   $\displaystyle f(x) = f(a) + \int_a^x f'(t)\,dt$
6. 6.25.6

   Integrating Rates of Change

   The integral of a rate of change gives the net change in the amount. Add the starting amount to get the final amount. For velocity, $\int v\,dt$ is displacement and $\int \lvert v \rvert\,dt$ is total distance.

   e.g. Tank has 10 gal, water flows in at $r(t) = 4t$ gal/min: after 3 min, $10 + \int_0^3 4t\,dt = 10 + 18 = 28$ gal

   $\displaystyle Q(b) = Q(a) + \int_a^b Q'(t)\,dt$
7. 6.25.7

   Integrating Density Functions

   Integrating a density (amount per unit length) over an interval gives the total amount. For a density per unit area that depends on the distance $r$ from a center, multiply by the ring circumference $2\pi r$.

   e.g. A 2 m rod with density $\rho(x) = 3x^2$ kg/m has mass $\int_0^2 3x^2\,dx = 8$ kg

   $\displaystyle \text{total} = \int_a^b \rho(x)\,dx$

   $\displaystyle \text{radial: } \int_0^R 2\pi r\,\rho(r)\,dr$

6.26 Integration Using Substitution

1. 6.26.1

   Integrating Algebraic Functions Using Substitution

   Let $u$ be the inner expression whose derivative also appears as a factor. Rewrite everything in $u$, integrate, then substitute back.

   e.g. $\int 2x(x^2+1)^3\,dx$: $u = x^2+1$ gives $\tfrac14(x^2+1)^4 + C$

   $\displaystyle \int f(g(x))\,g'(x)\,dx$

   $\displaystyle = \int f(u)\,du, \quad u = g(x)$
2. 6.26.2

   Integrating Linear Rational Functions Using Substitution

   For a fraction with a linear denominator, let $u = ax + b$, so $dx = \tfrac1a\,du$. If the numerator is also linear, write it in terms of $u$ and split the fraction.

   e.g. $\int \frac{3}{2x+1}\,dx = \tfrac32 \ln\lvert 2x+1\rvert + C$

   $\displaystyle \int \frac{1}{ax+b}\,dx$

   $\displaystyle = \frac1a \ln\lvert ax+b\rvert + C$
3. 6.26.3

   Integration Using Substitution

   Choose $u$, find $du$, and rewrite the whole integrand in $u$ with no $x$ left. A missing constant factor can be fixed by multiplying and dividing. Always substitute back at the end.

   e.g. $\int x\sqrt{x^2+4}\,dx$: $u = x^2+4$, $du = 2x\,dx$, so $\tfrac12\int u^{1/2}\,du = \tfrac13(x^2+4)^{3/2} + C$

   $\displaystyle u = g(x)$

   $\displaystyle du = g'(x)\,dx$
4. 6.26.4

   Calculating Definite Integrals Using Substitution

   Change the limits to $u$-values when you substitute. Then evaluate in $u$ with no need to substitute back.

   e.g. $\int_0^1 2x(x^2+1)\,dx$: $u$ runs from $1$ to $2$, so $\int_1^2 u\,du = 2 - \tfrac12 = \tfrac32$

   $\displaystyle \int_a^b f(g(x))\,g'(x)\,dx$

   $\displaystyle = \int_{g(a)}^{g(b)} f(u)\,du$
5. 6.26.5

   Further Integration of Algebraic Functions Using Substitution

   If an extra $x$ is left over after substituting, solve $u = g(x)$ for $x$ and replace it too. Then expand and integrate term by term.

   e.g. $\int x\sqrt{x+1}\,dx$: $u = x+1$, $x = u - 1$, so $\int (u^{3/2} - u^{1/2})\,du = \tfrac25 (x+1)^{5/2} - \tfrac23 (x+1)^{3/2} + C$

6. 6.26.6

   Integrating Exponential Functions Using Linear Substitution

   With a linear exponent, integrate as usual and divide by the coefficient of $x$.

   e.g. $\int e^{3x-1}\,dx = \tfrac13 e^{3x-1} + C$

   $\displaystyle \int e^{ax+b}\,dx = \frac1a e^{ax+b} + C$
7. 6.26.7

   Integrating Exponential Functions Using Substitution

   Let $u$ be the exponent. The derivative of the exponent must appear as a factor, up to a constant.

   e.g. $\int x e^{x^2}\,dx$: $u = x^2$, $du = 2x\,dx$, so the answer is $\tfrac12 e^{x^2} + C$

   $\displaystyle \int e^{g(x)}\,g'(x)\,dx = e^{g(x)} + C$
8. 6.26.8

   Integrating Trigonometric Functions Using Substitution

   Let $u$ be the angle, or let $u$ be $\sin x$ or $\cos x$ when its derivative is a factor. Watch the sign: $\int \sin u\,du = -\cos u + C$.

   e.g. $\int \sin^2 x\cos x\,dx$: $u = \sin x$ gives $\tfrac13\sin^3 x + C$

   $\displaystyle \int \cos(ax+b)\,dx$

   $\displaystyle = \frac1a \sin(ax+b) + C$

   $\displaystyle \int \sin(ax+b)\,dx$

   $\displaystyle = -\frac1a \cos(ax+b) + C$
9. 6.26.9

   Integrating Logarithmic Functions Using Substitution

   When $\ln x$ appears with a factor of $\tfrac1x$, let $u = \ln x$. A fraction whose numerator is the derivative of its denominator integrates to a natural log.

   e.g. $\int \frac{\ln x}{x}\,dx = \tfrac12(\ln x)^2 + C$

   $\displaystyle u = \ln x \Rightarrow du = \frac1x\,dx$

   $\displaystyle \int \frac{g'(x)}{g(x)}\,dx$

   $\displaystyle = \ln\lvert g(x)\rvert + C$
10. 6.26.10

    Integration by Substitution With Inverse Trigonometric Functions

    Match the integrand to an inverse trig form, using substitution to make the variable part a perfect square. Here $a > 0$.

    e.g. $\int \frac{dx}{1 + 4x^2}$: $u = 2x$ gives $\tfrac12\arctan(2x) + C$

    $\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$

    $\displaystyle \int \frac{dx}{a^2 + x^2}$

    $\displaystyle = \frac1a \arctan\frac{x}{a} + C$

6.27 Integration Using Trigonometric Identities

1. 6.27.1

   Integration Using Basic Trigonometric Identities

   Rewrite the integrand with quotient and reciprocal identities until it matches a known antiderivative.

   e.g. $\int \frac{\sin x}{\cos^2 x}\,dx = \int \sec x\tan x\,dx = \sec x + C$

   $\displaystyle \tan x = \frac{\sin x}{\cos x}$

   $\displaystyle \sec x = \frac{1}{\cos x},\quad \csc x = \frac{1}{\sin x}$
2. 6.27.2

   Integration Using the Pythagorean Identities

   Replace a square you cannot integrate with one you can, or split off one factor so substitution works.

   e.g. $\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C$

   $\displaystyle \sin^2 x + \cos^2 x = 1$

   $\displaystyle 1 + \tan^2 x = \sec^2 x$
3. 6.27.3

   Integration Using the Double-Angle Formulas

   To integrate $\sin^2 x$ or $\cos^2 x$, lower the power with these identities first.

   e.g. $\int \sin^2 x\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$

   $\displaystyle \sin^2 x = \frac{1 - \cos 2x}{2}$

   $\displaystyle \cos^2 x = \frac{1 + \cos 2x}{2}$

   $\displaystyle \sin x\cos x = \frac12 \sin 2x$

6.28 Special Techniques for Integration

1. 6.28.1

   Integrating Functions Using Polynomial Division

   If the numerator's degree is at least the denominator's, divide first. Integrate the quotient and the remainder fraction separately.

   e.g. $\int \frac{x^2+1}{x+1}\,dx = \int \left(x - 1 + \tfrac{2}{x+1}\right)dx = \tfrac{x^2}{2} - x + 2\ln\lvert x+1\rvert + C$

   $\displaystyle \frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$
2. 6.28.2

   Integrating Functions by Completing the Square

   For a quadratic denominator that does not factor, complete the square, then use the arctan form with $u = x + \tfrac b2$.

   e.g. $\int \frac{dx}{x^2 + 2x + 5} = \int \frac{dx}{(x+1)^2 + 4} = \tfrac12\arctan\tfrac{x+1}{2} + C$

   $\displaystyle x^2 + bx + c$

   $\displaystyle = \left(x + \tfrac b2\right)^2 + c - \tfrac{b^2}{4}$

   $\displaystyle \int \frac{du}{u^2 + a^2}$

   $\displaystyle = \frac1a \arctan\frac{u}{a} + C$

7

Unit 7 18 topics

Differential Equations

Introduction to Differential Equations, Qualitative Techniques for Differential Equations, and Slope Fields.

7.29 Introduction to Differential Equations

1. 7.29.1

   Introduction to Differential Equations

   A differential equation relates a function to its derivatives. Its order is the highest derivative that appears. A solution is a function, not a number.

   e.g. $\frac{dy}{dx} = 2x$ is first order; $y'' + y = 0$ is second order

2. 7.29.2

   Verifying Solutions of Differential Equations

   Compute the needed derivatives of the proposed function and substitute them into the equation. It is a solution only if both sides are equal for every $x$.

   e.g. $y = e^{3x}$ solves $y' = 3y$, since $y' = 3e^{3x} = 3y$

3. 7.29.3

   Solving First-Order ODEs Using Direct Integration

   If the derivative depends only on $x$, integrate both sides. Keep the $+ C$: it gives a whole family of solutions.

   e.g. $y' = 6x^2 \Rightarrow y = 2x^3 + C$

   $\displaystyle \frac{dy}{dx} = f(x)$

   $\displaystyle \Rightarrow\; y = \int f(x)\,dx + C$
4. 7.29.4

   Solving First-Order ODEs Using Separation of Variables

   Move all $y$ terms (with $dy$) to one side and all $x$ terms (with $dx$) to the other, then integrate both sides. One $+ C$ on the $x$ side is enough. Solve for $y$ if you can.

   e.g. $\frac{dy}{dx} = xy$: $\ln\lvert y\rvert = \tfrac{x^2}{2} + C$, so $y = Ae^{x^2/2}$

   $\displaystyle \frac{dy}{dx} = g(x)\,h(y)$

   $\displaystyle \Rightarrow\; \int \frac{dy}{h(y)} = \int g(x)\,dx$
5. 7.29.5

   Solving First-Order IVPs Using Separation of Variables

   Separate and integrate, then plug in the initial condition to find $C$. Use it also to pick the right sign when you take a square root.

   e.g. $\frac{dy}{dx} = \frac{x}{y}$, $y(0) = 2$: $\tfrac{y^2}{2} = \tfrac{x^2}{2} + 2$, so $y = \sqrt{x^2 + 4}$

6. 7.29.6

   Modeling With First-Order ODEs

   Translate the words into a rate equation. "Proportional to" means multiply by a constant $k$. Use a negative sign for a quantity that is decreasing.

   e.g. Water drains at a rate proportional to the square root of the depth $h$: $\frac{dh}{dt} = -k\sqrt{h}$

   $\displaystyle \frac{dy}{dt} = ky \quad \text{(proportional to } y\text{)}$
7. 7.29.7

   Further Modeling With First-Order ODEs

   In Newton's law of cooling, the rate of temperature change is proportional to the gap between the object and its surroundings. For mixing or flow problems, the net rate is rate in minus rate out.

   e.g. Coffee at 180°F in a 70°F room: $\frac{dT}{dt} = -k(T - 70)$ with $k > 0$

   $\displaystyle \frac{dT}{dt} = -k(T - T_s)$

   $\displaystyle \frac{dA}{dt} = \text{rate in} - \text{rate out}$
8. 7.29.8

   Exponential Growth and Decay Models With First-Order ODEs

   If the rate is proportional to the amount, the solution is exponential. $k > 0$ means growth and $k < 0$ means decay. $y_0$ is the amount at $t = 0$.

   e.g. $\frac{dy}{dt} = 0.05y$, $y(0) = 200$: $y = 200e^{0.05t}$

   $\displaystyle \frac{dy}{dt} = ky \;\Rightarrow\; y = y_0 e^{kt}$
9. 7.29.9

   Exponential Growth and Decay Models With First-Order ODEs: Calculating Unknown Times

   Set the model equal to the target amount, divide by $y_0$, and take ln of both sides.

   e.g. $200e^{0.05t} = 400 \Rightarrow t = \tfrac{\ln 2}{0.05} \approx 13.9$

   $\displaystyle y_0 e^{kt} = y$

   $\displaystyle \Rightarrow\; t = \frac1k \ln\frac{y}{y_0}$
10. 7.29.10

    Exponential Growth and Decay Models With First-Order ODEs: Half-Life Problems

    The half-life $T$ is the time for the amount to halve. It fixes the decay constant $k$.

    e.g. 64 g with a 5-day half-life: after 15 days, $64\left(\tfrac12\right)^3 = 8$ g

    $\displaystyle k = -\frac{\ln 2}{T}$

    $\displaystyle y = y_0 e^{kt} = y_0\left(\tfrac12\right)^{t/T}$

7.30 Qualitative Techniques for Differential Equations

1. 7.30.1

   Qualitative Analysis of First-Order ODEs

   Read behavior from $\frac{dy}{dx}$ without solving. Where it is positive, solutions increase; where negative, they decrease. Differentiate $\frac{dy}{dx}$ implicitly to get $\frac{d^2y}{dx^2}$ for concavity.

   e.g. $\frac{dy}{dx} = y(3 - y)$: solutions increase for $0 < y < 3$ and decrease for $y > 3$ or $y < 0$

   $\displaystyle \frac{dy}{dx} > 0 \Rightarrow y \text{ increasing}$

   $\displaystyle \frac{d^2y}{dx^2} > 0 \Rightarrow \text{concave up}$
2. 7.30.2

   Equilibrium Solutions of First-Order ODEs

   An equilibrium is a constant solution $y = c$, where $\frac{dy}{dx} = 0$ for every $x$. It is stable if nearby solutions move toward it and unstable if they move away.

   e.g. $\frac{dy}{dx} = y(3 - y)$: $y = 0$ is unstable and $y = 3$ is stable

   $\displaystyle \frac{dy}{dx} = f(y),\; f(c) = 0$

   $\displaystyle \Rightarrow\; y = c \text{ is an equilibrium}$

7.31 Slope Fields

1. 7.31.1

   Slope Fields for Directly Integrable Differential Equations

   For $\frac{dy}{dx} = f(x)$, the slope depends only on $x$. All segments in the same vertical column are parallel, and solution curves are vertical shifts of each other.

   e.g. $\frac{dy}{dx} = 2x$: slope $0$ along $x = 0$, slope $2$ along $x = 1$; solutions are $y = x^2 + C$

2. 7.31.2

   Slope Fields for Autonomous Differential Equations

   For $\frac{dy}{dx} = f(y)$, the slope depends only on $y$. All segments in the same horizontal row are parallel, and solution curves are horizontal shifts of each other. Rows where $f(y) = 0$ are flat.

   e.g. $\frac{dy}{dx} = y - 1$: slope $0$ along $y = 1$, slope $1$ along $y = 2$, slope $-1$ along $y = 0$

3. 7.31.3

   Slope Fields for Nonautonomous Differential Equations

   For $\frac{dy}{dx} = f(x, y)$, plug each grid point into $f$ to get its slope. The curve where $f(x, y) = 0$ has horizontal segments.

   e.g. $\frac{dy}{dx} = x - y$: slope $0$ at $(1, 1)$, $2$ at $(2, 0)$, $-1$ at $(0, 1)$; flat along $y = x$

4. 7.31.4

   Analyzing Slope Fields for Directly Integrable Differential Equations

   If segments in each vertical column are parallel, the equation has the form $\frac{dy}{dx} = f(x)$. Note where slopes are zero, positive, and negative to identify $f$.

   e.g. Slope $0$ on the $y$-axis, positive for $x > 0$, negative for $x < 0$: fits $\frac{dy}{dx} = x$

5. 7.31.5

   Analyzing Slope Fields for Autonomous Differential Equations

   If segments in each horizontal row are parallel, the equation has the form $\frac{dy}{dx} = f(y)$. Flat rows are equilibria. Arrows pointing toward a flat row mark it stable.

   e.g. Flat along $y = 2$, rising below it, falling above it: fits $\frac{dy}{dx} = 2 - y$, and solutions approach $y = 2$

6. 7.31.6

   Analyzing Slope Fields for Nonautonomous Differential Equations

   When slopes change in both directions, test a few points in each candidate equation. Look first for the line or curve of zero slopes and the sign of the slope on each side of it.

   e.g. Flat along $y = -x$ and rising above it: fits $\frac{dy}{dx} = x + y$

8

Unit 8 18 topics

Applications of Integration

Applications of Integration, Connecting Position, Velocity and Acceleration Using Integrals, Volumes of Solids With Known Cross Sections, and Volumes of Revolution.

8.32 Applications of Integration

1. 8.32.1

   The Average Value of a Function

   The average value of $f$ on $[a, b]$ is its integral divided by the length of the interval.

   e.g. $f(x) = x^2$ on $[0, 3]$: $\tfrac13\int_0^3 x^2\,dx = \tfrac13 \cdot 9 = 3$

   $\displaystyle f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$
2. 8.32.2

   The Area Between Curves Expressed as Functions of X

   Integrate top minus bottom with respect to $x$. The limits are usually the $x$-values where the curves intersect.

   e.g. $y = x$ and $y = x^2$ on $[0, 1]$: $\int_0^1 (x - x^2)\,dx = \tfrac12 - \tfrac13 = \tfrac16$

   $\displaystyle A = \int_a^b \left[f(x) - g(x)\right]dx$

   $\displaystyle f(x) \ge g(x) \text{ on } [a, b]$
3. 8.32.3

   The Area Between Curves Expressed as Functions of Y

   When the curves are easier to write as $x$ in terms of $y$, integrate right minus left with respect to $y$. The limits are $y$-values.

   e.g. $x = y^2$ and $x = y + 2$ meet at $y = -1, 2$: $\int_{-1}^{2} (y + 2 - y^2)\,dy = \tfrac92$

   $\displaystyle A = \int_c^d \left[x_R(y) - x_L(y)\right]dy$
4. 8.32.4

   Finding Areas Between Curves that Intersect at More Than Two Points

   Find every intersection. Between each pair, check which curve is on top, then add the separate areas so that no part cancels.

   e.g. $y = x^3$ and $y = x$ meet at $x = -1, 0, 1$: $\int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx = \tfrac14 + \tfrac14 = \tfrac12$

   $\displaystyle A = \int_a^b \lvert f(x) - g(x)\rvert\,dx$

8.33 Connecting Position, Velocity and Acceleration Using Integrals

1. 8.33.1

   Calculating Velocity Using Integration

   Velocity is an antiderivative of acceleration. Use the initial velocity to find $C$.

   e.g. $a(t) = 6t$, $v(0) = 2$: $v(t) = 3t^2 + 2$

   $\displaystyle v(t) = \int a(t)\,dt$

   $\displaystyle v(b) = v(a) + \int_a^b a(t)\,dt$
2. 8.33.2

   Determining Characteristics of Moving Objects Using Integration

   Integrate to get velocity or position, then read the motion. At rest when $v = 0$. Moving right when $v > 0$ and left when $v < 0$. Speeding up when $v$ and $a$ have the same sign, slowing down when opposite.

   e.g. $a(t) = 2$, $v(0) = -4$: $v(t) = 2t - 4$, at rest at $t = 2$, moving left and slowing down before that

   $\displaystyle \text{speeding up: } v, a \text{ same sign}$

   $\displaystyle \text{slowing down: opposite signs}$
3. 8.33.3

   Calculating the Position Function of a Particle Using Integration

   Position is an antiderivative of velocity. Use the initial position to find $C$.

   e.g. $v(t) = 4t - 3$, $s(0) = 5$: $s(t) = 2t^2 - 3t + 5$

   $\displaystyle s(t) = \int v(t)\,dt$

   $\displaystyle s(b) = s(a) + \int_a^b v(t)\,dt$
4. 8.33.4

   Calculating the Displacement of a Particle Using Integration

   Displacement is the net change in position: the integral of velocity. It can be negative or zero, because motion in opposite directions cancels.

   e.g. $v(t) = t - 2$ on $[0, 4]$: $\int_0^4 (t - 2)\,dt = 8 - 8 = 0$

   $\displaystyle \int_a^b v(t)\,dt = s(b) - s(a)$
5. 8.33.5

   Calculating the Total Distance Traveled by a Particle

   Total distance is the integral of speed, which is never negative. Split the interval where $v$ changes sign and add the absolute values.

   e.g. $v(t) = t - 2$ on $[0, 4]$: $\int_0^2 (2 - t)\,dt + \int_2^4 (t - 2)\,dt = 2 + 2 = 4$

   $\displaystyle \text{distance} = \int_a^b \lvert v(t)\rvert\,dt$
6. 8.33.6

   Average Position, Velocity, and Acceleration

   Each is the average value of that function on $[a, b]$. Average velocity is also change in position over time, and average acceleration is change in velocity over time.

   e.g. $v(t) = 3t^2$ on $[0, 2]$: average velocity $\tfrac12\int_0^2 3t^2\,dt = \tfrac12 \cdot 8 = 4$

   $\displaystyle \bar v = \frac{1}{b-a}\int_a^b v(t)\,dt$

   $\displaystyle = \frac{s(b) - s(a)}{b - a}$

   $\displaystyle \bar a = \frac{v(b) - v(a)}{b - a}$

   $\displaystyle \bar s = \frac{1}{b-a}\int_a^b s(t)\,dt$

8.34 Volumes of Solids With Known Cross Sections

1. 8.34.1

   Volumes of Solids with Square Cross Sections

   Slice perpendicular to the axis. Each slice is a square whose side $s$ is the length across the base region at that point. Integrate the area of the square.

   e.g. Base under $y = \sqrt{x}$ on $[0, 4]$, squares perpendicular to the $x$-axis: $s = \sqrt{x}$, $V = \int_0^4 x\,dx = 8$

   $\displaystyle A(x) = [s(x)]^2$

   $\displaystyle V = \int_a^b [s(x)]^2\,dx$
2. 8.34.2

   Volumes of Solids with Rectangular Cross Sections

   One side $s$ of each rectangle lies in the base region. The other side $h$ is given, often a constant or a multiple of $s$. Integrate base times height.

   e.g. Base under $y = x$ on $[0, 2]$, height 3 times the base: $A = 3x^2$, $V = \int_0^2 3x^2\,dx = 8$

   $\displaystyle A(x) = s(x)\,h(x)$

   $\displaystyle V = \int_a^b s(x)\,h(x)\,dx$
3. 8.34.3

   Volumes of Solids with Triangular Cross Sections

   Find the side $s$ that lies in the base, then use the area formula for the stated triangle. Read carefully whether $s$ is a leg or the hypotenuse.

   e.g. Base under $y = 2x$ on $[0, 3]$, equilateral triangles: $A = \tfrac{\sqrt3}{4}(2x)^2 = \sqrt3\,x^2$, $V = \sqrt3\int_0^3 x^2\,dx = 9\sqrt3$

   $\displaystyle A = \tfrac{\sqrt3}{4}s^2 \;\;\text{equilateral}$

   $\displaystyle A = \tfrac12 s^2 \;\;\text{isos. right, leg } s$

   $\displaystyle A = \tfrac14 s^2 \;\;\text{isos. right, hyp. } s$

   $\displaystyle V = \int_a^b A(x)\,dx$
4. 8.34.4

   Volumes of Solids with Circular Cross Sections

   The length $s$ across the base is the diameter, so the radius is $s/2$. For semicircles, take half the circle's area.

   e.g. Diameter $s = x$ on $[0, 2]$: $V = \int_0^2 \tfrac{\pi}{4}x^2\,dx = \tfrac{\pi}{4}\cdot\tfrac83 = \tfrac{2\pi}{3}$

   $\displaystyle A = \pi\left(\tfrac{s}{2}\right)^2 = \tfrac{\pi}{4}s^2$

   $\displaystyle A_{\text{semicircle}} = \tfrac{\pi}{8}s^2$

   $\displaystyle V = \int_a^b A(x)\,dx$

8.35 Volumes of Revolution

1. 8.35.1

   Volumes of Revolution Using the Disc Method: Rotation About the Coordinate Axes

   Use when the region touches the axis of rotation, so each slice is a solid disc. Rotating about the $x$-axis, integrate in $x$. Rotating about the $y$-axis, write the curve as $x = g(y)$ and integrate in $y$.

   e.g. $y = \sqrt{x}$, $0 \le x \le 4$, about the $x$-axis: $V = \pi\int_0^4 x\,dx = 8\pi$

   $\displaystyle V = \pi\int_a^b [f(x)]^2\,dx$

   $\displaystyle V = \pi\int_c^d [g(y)]^2\,dy$
2. 8.35.2

   Volumes of Revolution Using the Disc Method: Rotation About Other Axes

   The radius is the distance from the curve to the axis, not the curve's height. About $y = k$, $R = \lvert f(x) - k\rvert$ and integrate in $x$. About $x = h$, $R = \lvert g(y) - h\rvert$ and integrate in $y$.

   e.g. Region between $y = x^2$ and $y = 1$, about $y = 1$: $R = 1 - x^2$, $V = \pi\int_{-1}^{1}(1 - x^2)^2\,dx = \tfrac{16\pi}{15}$

   $\displaystyle V = \pi\int_a^b [R(x)]^2\,dx$

   $\displaystyle R(x) = \lvert f(x) - k\rvert$
3. 8.35.3

   Volumes of Revolution Using the Washer Method: Rotation About the Coordinate Axes

   Use when there is a gap between the region and the axis. Each slice is a washer with outer radius $R$ and inner radius $r$. Square each radius separately: $R^2 - r^2$, not $(R - r)^2$.

   e.g. Between $y = x$ and $y = x^2$ on $[0, 1]$, about the $x$-axis: $V = \pi\int_0^1 (x^2 - x^4)\,dx = \pi\left(\tfrac13 - \tfrac15\right) = \tfrac{2\pi}{15}$

   $\displaystyle V = \pi\int_a^b \left(R^2 - r^2\right)dx$
4. 8.35.4

   Volumes of Revolution Using the Washer Method: Rotation About Other Axes

   Measure both radii from the axis of rotation. $R$ goes to the curve farther from the axis, $r$ to the nearer one. For a vertical axis $x = h$, use $x$ in terms of $y$ and integrate in $y$.

   e.g. Between $y = x$ and $y = x^2$ on $[0, 1]$, about $y = -1$: $R = x + 1$, $r = x^2 + 1$, $V = \pi\int_0^1 (2x - x^2 - x^4)\,dx = \tfrac{7\pi}{15}$

   $\displaystyle R = \lvert y_{\text{far}} - k\rvert$

   $\displaystyle r = \lvert y_{\text{near}} - k\rvert$

   $\displaystyle V = \pi\int_a^b \left(R^2 - r^2\right)dx$

9

Unit 9 7 topics

Applications of Technology

Using Graphing Calculators.

9.36 Using Graphing Calculators

1. 9.36.1

   Evaluating Expressions Using a Graphing Calculator

   Type the expression with full parentheses around numerators, denominators, and exponents. Use radian mode for calculus. On the AP exam, report decimals correct to three places.

   e.g. $e^{0.5}\sin 2 \approx 1.6487 \times 0.9093 \approx 1.499$ (radians)

2. 9.36.2

   Finding Roots of Functions Using a Graphing Calculator

   Graph $y = f(x)$ and find where it crosses the $x$-axis. On a TI-84 use CALC, zero (left bound, right bound, guess). In Desmos, click the $x$-intercept.

   e.g. $x^3 - x - 1 = 0$: one real root, $x \approx 1.325$

3. 9.36.3

   Finding Intersections of Functions Using a Graphing Calculator

   Graph both functions and find where they cross. On a TI-84 use CALC, intersect. In Desmos, click the crossing point. You can also find the roots of $f(x) - g(x)$.

   e.g. $\cos x = x$ at $x \approx 0.739$

4. 9.36.4

   Finding Extrema of Functions Using a Graphing Calculator

   Graph the function and use CALC, maximum or minimum on a TI-84, or click the turning point in Desmos. On a closed interval, also check the endpoint values.

   e.g. $f(x) = xe^{-x}$: local max at $(1, 0.368)$

5. 9.36.5

   Finding Derivatives Using a Graphing Calculator

   The calculator gives the derivative's value at one point, not a formula. On a TI-84 use nDeriv( or the $\tfrac{d}{dx}$ template. In Desmos, define $f(x)$ and type $f'(a)$.

   e.g. $f(x) = x^x$ at $x = 2$: $f'(2) = 4(\ln 2 + 1) \approx 6.773$

6. 9.36.6

   Finding Definite Integrals Using a Graphing Calculator

   Use fnInt( or the $\int$ template on a TI-84, or type the integral in Desmos. Use it for integrals with no simple antiderivative. The result is a signed area.

   e.g. $\int_0^1 e^{x^2}\,dx \approx 1.463$

7. 9.36.7

   Exploring Functions Using Technology

   Use the graph, a table of values, and zoom to see a function's behavior. Graph $f'$ to see where $f$ increases ($f' > 0$) or decreases ($f' < 0$). Graphs suggest answers. Calculus confirms them.

   e.g. Graph $y = \tfrac{\sin x}{x}$: values approach 1 near $x = 0$, suggesting $\lim_{x \to 0} \tfrac{\sin x}{x} = 1$

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