Exercise 1: The general antiderivative: read the table backwards, then differentiate

A function $F$ is an antiderivative of $f$ on an interval $I$ when $F'(x) = f(x)$ for every $x$ in $I$ . If $F$ is one antiderivative, the most general antiderivative of $f$ on $I$ is $F(x) + C$ , with $C$ an arbitrary constant: two antiderivatives on an INTERVAL differ by a constant, a consequence of the Mean Value Theorem.

Every line of the table of antiderivatives is a line of the table of derivatives read from right to left, and so every answer can be checked in one line by differentiating it. That check is part of the answer on this set, and it is the only protection against a missing factor.

a) Find the most general antiderivative of $f(x) = 8x^3 - 6\sqrt{x} + \frac{3}{x^4}$ , and state the interval on which it is valid.
b) Same question for $g(x) = \frac{5}{x} + 4e^x - \frac{2}{1 + x^2}$ .
c) Same question for $h(\theta) = 3\cos\theta - 2\sec^2\theta + \sec\theta\tan\theta$ .
d) Same question for $k(x) = \frac{3}{\sqrt{1 - x^2}} + 10^x$ on $(-1, 1)$ . A student answers $3\arcsin x + \frac{10^{x+1}}{x+1} + C$ : find the error by differentiating.
e) The table has no line for $\ln x$ . Decide, by differentiating, which of $F_1(x) = x\ln x - x + 7$ , $F_2(x) = \frac{(\ln x)^2}{2}$ and $F_3(x) = x\ln x$ are antiderivatives of $\ln x$ on $(0, \infty)$ .

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Answers

a) $F(x) = 2x^4 - 4x^{3/2} - x^{-3} + C$ on $(0, \infty)$
b) $G(x) = 5\ln|x| + 4e^x - 2\arctan x + C$ , one constant on $(-\infty, 0)$ and one on $(0, \infty)$
c) $H(\theta) = 3\sin\theta - 2\tan\theta + \sec\theta + C$ on each interval where $\cos\theta \ne 0$
d) $K(x) = 3\arcsin x + \frac{10^x}{\ln 10} + C$ ; the power rule does not apply to $10^x$
e) Only $F_1$ : $F_1'(x) = \ln x$ , $F_2'(x) = \frac{\ln x}{x}$ , $F_3'(x) = \ln x + 1$

a) Rewrite every term as a power of $x$ BEFORE reading the table: $f(x) = 8x^3 - 6x^{1/2} + 3x^{-4}$ . The power rule backwards, $x^n \mapsto \frac{x^{n+1}}{n+1}$ for $n \ne -1$ , gives $8 \cdot \frac{x^4}{4} = 2x^4$ , then $-6 \cdot \frac{x^{3/2}}{3/2} = -4x^{3/2}$ , then $3 \cdot \frac{x^{-3}}{-3} = -x^{-3}$ . So $F(x) = 2x^4 - 4x^{3/2} - x^{-3} + C$ . The domain of $f$ is $(0, \infty)$ , because of $\sqrt x$ and of $x^{-4}$ together: it is ONE interval, so one constant suffices. Check: $F'(x) = 8x^3 - 4 \cdot \frac{3}{2}x^{1/2} + 3x^{-4} = 8x^3 - 6\sqrt x + \frac{3}{x^4}$ . The classic slip is $\frac{3}{x^4} \mapsto \frac{3}{x^5} \cdot \frac{1}{5}$ , raising the exponent in the denominator: the exponent of $x^{-4}$ goes UP to $-3$ , and the differentiation catches the error at once.

b) $5 \cdot \frac{1}{x}$ comes from $5\ln|x|$ , $4e^x$ from $4e^x$ , and $-\frac{2}{1 + x^2}$ from $-2\arctan x$ , since $(\arctan x)' = \frac{1}{1 + x^2}$ . Check: $G'(x) = \frac{5}{x} + 4e^x - \frac{2}{1 + x^2}$ . The domain of $g$ is $x \ne 0$ , which is TWO intervals, $(-\infty, 0)$ and $(0, \infty)$ : the most general antiderivative carries one constant on each, $G(x) = 5\ln|x| + 4e^x - 2\arctan x + C_1$ for $x > 0$ and $+ C_2$ for $x < 0$ . Exercise 5 is devoted to this. Writing $\ln x$ without the absolute value gives an antiderivative on $(0, \infty)$ only, and loses half of the domain.

c) Read three lines of the derivative table backwards: $(\sin\theta)' = \cos\theta$ , $(\tan\theta)' = \sec^2\theta$ , $(\sec\theta)' = \sec\theta\tan\theta$ . So $H(\theta) = 3\sin\theta - 2\tan\theta + \sec\theta + C$ . Check: $H'(\theta) = 3\cos\theta - 2\sec^2\theta + \sec\theta\tan\theta$ . The signs are where the marks go: the antiderivative of $\cos$ is $+\sin$ , of $\sin$ is $-\cos$ , and a student who remembers only that a minus sign appears somewhere gets half of them wrong. The domain excludes the zeros of $\cos\theta$ , so the formula holds on each interval $\left(-\frac{\pi}{2} + k\pi, \frac{\pi}{2} + k\pi\right)$ , with its own constant.

d) $(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$ on $(-1, 1)$ , and $(10^x)' = 10^x \ln 10$ , so $10^x$ comes from $\frac{10^x}{\ln 10}$ : $K(x) = 3\arcsin x + \frac{10^x}{\ln 10} + C$ . Check: $K'(x) = \frac{3}{\sqrt{1 - x^2}} + \frac{10^x \ln 10}{\ln 10}$ . The student's term $\frac{10^{x+1}}{x+1}$ applies the power rule to an EXPONENTIAL: the variable is in the exponent, not in the base. Differentiating it by the quotient rule gives $\frac{10^{x+1}\ln 10 \cdot (x+1) - 10^{x+1}}{(x+1)^2}$ , which at $x = 0$ equals $10\ln 10 - 10 \approx 13$ , while $10^0 = 1$ . The power rule is for $x^n$ with a fixed exponent, never for $b^x$ .

e) $F_1'(x) = \left(1 \cdot \ln x + x \cdot \frac{1}{x}\right) - 1 + 0 = \ln x$ , by the product rule: $F_1$ IS an antiderivative of $\ln x$ , and the constant $7$ changes nothing. $F_2'(x) = \frac{2\ln x}{2} \cdot \frac{1}{x} = \frac{\ln x}{x}$ , by the chain rule with inner function $\ln x$ : not $\ln x$ . $F_3'(x) = \ln x + 1$ : off by the constant $1$ , so no. Only $F_1$ works, and the general antiderivative is $x\ln x - x + C$ . No line of the table produced it, and none will until MATH 141; but checking a PROPOSED antiderivative needs nothing more than the rules of differentiation, which is why a claimed antiderivative is never accepted without the check.

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Exercise 2: Rewrite before antidifferentiating: no product rule and no quotient rule backwards

The derivative table has a product rule, a quotient rule and a chain rule. The antiderivative table has NONE of them: the antiderivative of a product is not the product of the antiderivatives, and the antiderivative of a quotient is not the quotient. Before any table is read, the function is rewritten as a SUM of table entries, by expanding, by splitting a fraction over a single-term denominator, or by a long division.

One composition is allowed in MATH 140, the one with a LINEAR inside function: if $F' = f$ and $a \ne 0$ , then $\frac{1}{a}F(ax + b)$ is an antiderivative of $f(ax + b)$ . It is guessed, then proved by the chain rule, never assumed.

a) Find the most general antiderivative of $p(x) = (x^2 + 1)(2x - 3)$ . A student answers $\left(\frac{x^3}{3} + x\right)(x^2 - 3x) + C$ : show that it is wrong by comparing derivatives at $x = 1$ .
b) Same question for $q(x) = \frac{x^4 - 3x^2 + 6}{x^3}$ .
c) Same question for $r(x) = \frac{x^2}{x^2 + 1}$ . Hint: add and subtract $1$ in the numerator.
d) Same question for $s(x) = \frac{x + 3}{x + 1}$ . Guess the antiderivative of $\frac{1}{x + 1}$ and verify it.
e) Prove the rule for $f(ax + b)$ stated above, then find antiderivatives of $(3x - 1)^4$ and of $e^{5 - 2x}$ . A student writes $\frac{(3x - 1)^5}{5}$ : what does its derivative give?

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a) $\frac{x^4}{2} - x^3 + x^2 - 3x + C$ ; at $x = 1$ the student's derivative is $-\frac{16}{3}$ , not $p(1) = -2$
b) $\frac{x^2}{2} - 3\ln|x| - \frac{3}{x^2} + C$ , one constant on each side of $0$
c) $x - \arctan x + C$
d) $x + 2\ln|x + 1| + C$ , one constant on each side of $-1$
e) $\frac{(3x - 1)^5}{15} + C$ and $-\frac{1}{2}e^{5 - 2x} + C$ ; the student's answer differentiates to $3(3x - 1)^4$

a) Expand first: $p(x) = 2x^3 - 3x^2 + 2x - 3$ , a sum of powers. Term by term, $P(x) = \frac{x^4}{2} - x^3 + x^2 - 3x + C$ , and $P'(x) = 2x^3 - 3x^2 + 2x - 3 = p(x)$ . The student multiplied the antiderivatives of the two factors, $\frac{x^3}{3} + x$ and $x^2 - 3x$ . By the product rule, the derivative of that product is $(x^2 + 1)(x^2 - 3x) + \left(\frac{x^3}{3} + x\right)(2x - 3)$ , which at $x = 1$ equals $2 \cdot (-2) + \frac{4}{3} \cdot (-1) = -\frac{16}{3}$ , while $p(1) = 2 \cdot (-1) = -2$ . One point of disagreement is enough to reject a candidate. The reason is structural: differentiating a product produces TWO terms, so undoing a product cannot give one.

b) Split over the single-term denominator: $q(x) = \frac{x^4}{x^3} - \frac{3x^2}{x^3} + \frac{6}{x^3} = x - \frac{3}{x} + 6x^{-3}$ . Now each term is in the table: $\frac{x^2}{2}$ , then $-3\ln|x|$ (the exponent $-1$ is the one case of the power rule that does NOT apply), then $6 \cdot \frac{x^{-2}}{-2} = -3x^{-2}$ . So $Q(x) = \frac{x^2}{2} - 3\ln|x| - \frac{3}{x^2} + C$ , with a constant on each of $(-\infty, 0)$ and $(0, \infty)$ . Check: $Q'(x) = x - \frac{3}{x} + 6x^{-3}$ . Dividing a sum by a single term is legal term by term; dividing BY a sum, $\frac{1}{x^2 + 1}$ , is not, and c) shows what to do then.

c) The denominator is a sum, so the fraction does not split. Make the numerator contain the denominator: $\frac{x^2}{x^2 + 1} = \frac{(x^2 + 1) - 1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$ , the result of a long division. Both pieces are in the table: $R(x) = x - \arctan x + C$ , valid on all of $\mathbb{R}$ , a single interval. Check: $R'(x) = 1 - \frac{1}{1 + x^2} = \frac{x^2}{1 + x^2}$ . A student who writes $\frac{x^3/3}{\arctan x}$ , the quotient of two antiderivatives, obtains a function whose derivative has nothing to do with $r$ : the quotient rule does not run backwards either.

d) Same gesture: $\frac{x + 3}{x + 1} = \frac{(x + 1) + 2}{x + 1} = 1 + \frac{2}{x + 1}$ . For $\frac{1}{x + 1}$ , guess $\ln|x + 1|$ and verify with the chain rule, inner function $u = x + 1$ with $u' = 1$ : $\frac{d}{dx}\ln|x + 1| = \frac{1}{x + 1} \cdot 1$ . So $S(x) = x + 2\ln|x + 1| + C$ , one constant on $(-\infty, -1)$ and one on $(-1, \infty)$ . Check: $S'(x) = 1 + \frac{2}{x + 1} = \frac{x + 3}{x + 1}$ . Here the inner derivative is $1$ and no factor appears; in e) it is not $1$ .

e) If $F' = f$ , the chain rule with inner function $u = ax + b$ , $u' = a$ , gives $\frac{d}{dx}\left[\frac{1}{a}F(ax + b)\right] = \frac{1}{a} \cdot f(ax + b) \cdot a = f(ax + b)$ , which proves the rule; $a \ne 0$ is needed to divide. For $(3x - 1)^4$ : $F(u) = \frac{u^5}{5}$ and $a = 3$ , so $\frac{(3x - 1)^5}{15} + C$ . For $e^{5 - 2x}$ : $a = -2$ , so $-\frac{1}{2}e^{5 - 2x} + C$ , and the check gives $-\frac{1}{2} \cdot e^{5 - 2x} \cdot (-2) = e^{5 - 2x}$ . The student's $\frac{(3x - 1)^5}{5}$ differentiates to $(3x - 1)^4 \cdot 3$ , three times too much: the factor $\frac{1}{a}$ is exactly what the chain rule will put back. The rule works only because $a$ is a CONSTANT; with $x^2 + 1$ inside, dividing by $2x$ fails, as Exercise 8 shows.

Exercise 3: Trigonometric functions: an identity first, then the table

The table knows six trigonometric antiderivatives, those of $\cos x$ , $\sin x$ , $\sec^2 x$ , $\csc^2 x$ , $\sec x\tan x$ and $\csc x\cot x$ . It knows neither $\cos^2 x$ , nor a quotient of trigonometric functions, nor a square of a sum. Each function below becomes a sum of table entries after ONE identity: $\sin^2 x + \cos^2 x = 1$ , $\cos 2x = \cos^2 x - \sin^2 x$ , the half-angle forms $\cos^2 x = \frac{1 + \cos 2x}{2}$ and $\sin^2 x = \frac{1 - \cos 2x}{2}$ , or a multiplication by a conjugate.

Work on an interval where every expression is defined, and check each answer by differentiating.

a) Find the most general antiderivative of $\sec x(\sec x + \tan x)$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .
b) Same question for $\frac{\cos 2x}{\cos x - \sin x}$ on $\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$ .
c) Same question for $\frac{1}{1 + \cos x}$ on $(0, \pi)$ . Hint: multiply numerator and denominator by $1 - \cos x$ .
d) Same question for $\cos^2 x$ on $\mathbb{R}$ .
e) A student writes that an antiderivative of $\sin^2 x$ is $\frac{\sin^3 x}{3}$ . Differentiate it to find the error, then give a correct antiderivative.

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Answers

a) $\tan x + \sec x + C$
b) $\sin x - \cos x + C$
c) $\csc x - \cot x + C$
d) $\frac{x}{2} + \frac{\sin 2x}{4} + C$
e) $\left(\frac{\sin^3 x}{3}\right)' = \sin^2 x\cos x$ ; correct: $\frac{x}{2} - \frac{\sin 2x}{4} + C$

a) Distribute: $\sec x(\sec x + \tan x) = \sec^2 x + \sec x\tan x$ , two lines of the table. So the answer is $\tan x + \sec x + C$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , where $\cos x \ne 0$ . Check: $(\tan x + \sec x)' = \sec^2 x + \sec x\tan x$ . Nothing here needs more than expanding a product, the gesture of Exercise 2 a): products are opened before the table is read, whatever the functions.

b) Factor the numerator with $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$ . On the given interval $\cos x - \sin x \ne 0$ (it vanishes only at $x = \frac{\pi}{4} + k\pi$ ), so the factor cancels: the function is $\cos x + \sin x$ , and its antiderivative is $\sin x - \cos x + C$ . Check: $(\sin x - \cos x)' = \cos x + \sin x$ . The cancellation is legitimate only where the factor is not zero, and naming the interval says exactly that. Without the identity, no rule of the table touches a quotient.

c) On $(0, \pi)$ , $1 + \cos x \ne 0$ and $\sin x \ne 0$ . Multiply by the conjugate: $\frac{1}{1 + \cos x} \cdot \frac{1 - \cos x}{1 - \cos x} = \frac{1 - \cos x}{1 - \cos^2 x} = \frac{1 - \cos x}{\sin^2 x}$ . Split over the single-term denominator: $\frac{1}{\sin^2 x} - \frac{\cos x}{\sin^2 x} = \csc^2 x - \csc x\cot x$ . Read backwards: $(-\cot x)' = \csc^2 x$ and $(-\csc x)' = \csc x\cot x$ , so the antiderivative is $-\cot x + \csc x + C$ . Check: $(\csc x - \cot x)' = -\csc x\cot x + \csc^2 x$ . At $x = \frac{\pi}{2}$ , the function is $\frac{1}{1 + 0} = 1$ and the check gives $-1 \cdot 0 + 1 = 1$ . The two minus signs of $(\cot x)'$ and $(\csc x)'$ are where this question is lost.

d) The half-angle identity turns a square into a first power: $\cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x$ . The second term has a linear inside function, $2x$ , so by the rule of Exercise 2 e) its antiderivative is $\frac{1}{2} \cdot \frac{\sin 2x}{2}$ . Hence $\frac{x}{2} + \frac{\sin 2x}{4} + C$ . Check: $\frac{1}{2} + \frac{2\cos 2x}{4} = \frac{1 + \cos 2x}{2} = \cos^2 x$ . Forgetting the $\frac{1}{2}$ of the linear rule gives $\frac{x}{2} + \frac{\sin 2x}{2}$ , whose derivative $\frac{1}{2} + \cos 2x$ reaches $\frac{3}{2}$ at $x = 0$ , more than the maximum $1$ of $\cos^2 x$ .

e) By the chain rule with inner function $\sin x$ : $\left(\frac{\sin^3 x}{3}\right)' = \frac{3\sin^2 x}{3} \cdot \cos x = \sin^2 x\cos x$ . The factor $\cos x$ , the derivative of the inside, is the proof that the power rule does not apply to a POWER OF A FUNCTION: it applies to a power of $x$ . At $x = \frac{\pi}{3}$ the student's derivative is $\frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$ , while $\sin^2\frac{\pi}{3} = \frac{3}{4}$ . The right way is the other half-angle identity, $\sin^2 x = \frac{1 - \cos 2x}{2}$ , which gives $\frac{x}{2} - \frac{\sin 2x}{4} + C$ . Check: $\frac{1}{2} - \frac{\cos 2x}{2} = \sin^2 x$ . Adding d) and e): the antiderivative of $\cos^2 x + \sin^2 x = 1$ is $\frac{x}{2} + \frac{x}{2} = x$ , as it must be.

Exercise 4: Initial value problems: one condition for each constant

An initial value problem gives a derivative and one or more values of the unknown function, and asks for THE function. The method never changes: antidifferentiate, writing the constant at the very line where it appears; then use each condition to find one constant; then check both the equation and the conditions.

Each antidifferentiation brings its own constant. A second derivative therefore produces two constants, $f(x) = G(x) + Cx + D$ , and needs two conditions.

a) Find $f$ if $f'(x) = 3x^2 - 4x + 1$ and $f(2) = 5$ .
b) Find $f$ if $f'(x) = \sqrt{x}\,(3 - x)$ for $x > 0$ and $f(4) = 0$ . Then give $f(1)$ .
c) Find $f$ if $f'(x) = \frac{1}{1 + x^2}$ and $f(1) = 0$ . Then give $f(-1)$ and $\lim_{x\to\infty} f(x)$ .
d) Find $f$ if $f''(x) = 6x + 2$ , $f(0) = 1$ and $f(1) = 4$ .
e) Find $f$ if $f''(x) = 12x^2 + 4e^{2x}$ , $f(0) = 0$ and $f'(0) = 1$ . A student writes $f'(x) = 4x^3 + 2e^{2x} + C$ , then $f(x) = x^4 + e^{2x} + C$ : find what is wrong.

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Answers

a) $f(x) = x^3 - 2x^2 + x + 3$
b) $f(x) = 2x^{3/2} - \frac{2}{5}x^{5/2} - \frac{16}{5}$ , and $f(1) = -\frac{8}{5}$
c) $f(x) = \arctan x - \frac{\pi}{4}$ ; $f(-1) = -\frac{\pi}{2}$ ; limit $\frac{\pi}{4}$
d) $f(x) = x^3 + x^2 + x + 1$
e) $f(x) = x^4 + e^{2x} - x - 1$ ; the student lost the term $Cx$ and used one constant for two

a) The general antiderivative is $f(x) = x^3 - 2x^2 + x + C$ , the constant written at once. The condition gives $f(2) = 8 - 8 + 2 + C = 2 + C = 5$ , so $C = 3$ and $f(x) = x^3 - 2x^2 + x + 3$ . Check: $f'(x) = 3x^2 - 4x + 1$ and $f(2) = 8 - 8 + 2 + 3 = 5$ . The student who writes $f(x) = x^3 - 2x^2 + x$ and then notices that $f(2) = 2 \ne 5$ has nowhere to go: without $C$ , the condition cannot be satisfied, and the question loses all its marks at the very first line.

b) Expand before reading the table: $f'(x) = 3x^{1/2} - x^{3/2}$ . Then $f(x) = 3 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C = 2x^{3/2} - \frac{2}{5}x^{5/2} + C$ . With $4^{3/2} = 8$ and $4^{5/2} = 32$ : $f(4) = 16 - \frac{64}{5} + C = \frac{16}{5} + C = 0$ , so $C = -\frac{16}{5}$ . Then $f(1) = 2 - \frac{2}{5} - \frac{16}{5} = -\frac{8}{5}$ . Check: $f'(x) = 3x^{1/2} - x^{3/2} = \sqrt{x}\,(3 - x)$ . Evaluating $4^{5/2}$ as $(\sqrt 4)^5 = 2^5$ , root first, keeps the numbers small without a calculator.

c) $f(x) = \arctan x + C$ on $\mathbb{R}$ , one interval, one constant. $f(1) = \frac{\pi}{4} + C = 0$ gives $C = -\frac{\pi}{4}$ . Then $f(-1) = -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2}$ , since $\arctan$ is odd, and $\lim_{x\to\infty} f(x) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$ . The answers are exact: $C = -\frac{\pi}{4}$ , never $-0.785$ . The limit shows something the formula makes plain: this $f$ increases for ever, since $f' > 0$ , and yet stays below $\frac{\pi}{4}$ .

d) First antiderivative: $f'(x) = 3x^2 + 2x + C$ . Second: $f(x) = x^3 + x^2 + Cx + D$ . Two constants, two conditions: $f(0) = D = 1$ , then $f(1) = 1 + 1 + C + 1 = 4$ , so $C = 1$ . Hence $f(x) = x^3 + x^2 + x + 1$ . Check: $f'(x) = 3x^2 + 2x + 1$ , $f''(x) = 6x + 2$ , $f(0) = 1$ , $f(1) = 4$ . The two conditions here are both on $f$ , at two different points: nothing requires one of them to be on $f'$ , only that together they fix both constants.

e) $f'(x) = 4x^3 + 2e^{2x} + C_1$ , using the linear rule for $e^{2x}$ . $f'(0) = 2 + C_1 = 1$ , so $C_1 = -1$ . Then $f(x) = x^4 + e^{2x} - x + C_2$ , and $f(0) = 1 + C_2 = 0$ , so $C_2 = -1$ : $f(x) = x^4 + e^{2x} - x - 1$ . Check: $f'(x) = 4x^3 + 2e^{2x} - 1$ , $f''(x) = 12x^2 + 4e^{2x}$ , $f'(0) = 1$ , $f(0) = 0$ . The student made two errors in one line: the antiderivative of the constant $C$ is $Cx$ , not $C$ , and a second antidifferentiation brings a NEW constant. His $f(x) = x^4 + e^{2x} + C$ has $f'(0) = 2$ whatever $C$ is, so the condition $f'(0) = 1$ can never hold. Name the constants $C_1$ , $C_2$ from the start and find $C_1$ BEFORE the second antidifferentiation.

Exercise 5: One constant per interval: when the domain has a gap

The statement two antiderivatives differ by a constant comes from the Mean Value Theorem: if $F' = G'$ on an INTERVAL, then $F - G$ has zero derivative there, so it is constant. On a domain made of several intervals, the theorem applies on each piece separately, and the constant may change from one piece to the next.

The figure shows $y = \ln|x|$ in blue and another function $G$ in orange, defined for $x \ne 0$ : on each side of $0$ , the orange curve is the blue one moved vertically. The red points are on the graph of $G$ .

a) Prove that $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for every $x \ne 0$ , treating $x > 0$ and $x < 0$ separately.
b) Read $G$ on the figure and write its formula. Show that $G'(x) = \frac{1}{x}$ for every $x \ne 0$ , and yet that $G - \ln|x|$ is not constant. Why is there no contradiction? Write the most general antiderivative of $\frac{1}{x}$ .
c) Find $f$ such that $f'(x) = \frac{1}{x^2}$ for $x \ne 0$ , $f(1) = 0$ and $f(-1) = 0$ . A classmate claims that no such $f$ exists, because $-\frac{1}{x} + 1$ gives $f(-1) = 2$ : answer him.
d) Find $f$ on $\left(-\frac{\pi}{2}, \frac{3\pi}{2}\right)$ minus the point $\frac{\pi}{2}$ such that $f'(x) = \sec^2 x$ , $f(0) = 0$ and $f(\pi) = 1$ .
e) Find $f$ such that $f'(x) = \frac{1}{x - 2}$ for $x \ne 2$ , $f(3) = 0$ and $f(0) = 1$ .

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Answers

a) For $x > 0$ , $(\ln x)' = \frac{1}{x}$ ; for $x < 0$ , $(\ln(-x))' = \frac{-1}{-x} = \frac{1}{x}$
b) $G(x) = \ln x + 1$ for $x > 0$ , $\ln(-x) - 1$ for $x < 0$ ; $G - \ln|x|$ is $1$ on the right, $-1$ on the left; general: $\ln|x| + C_1$ ( $x > 0$ ), $\ln|x| + C_2$ ( $x < 0$ )
c) $f(x) = 1 - \frac{1}{x}$ for $x > 0$ and $f(x) = -1 - \frac{1}{x}$ for $x < 0$
d) $f(x) = \tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $f(x) = \tan x + 1$ on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$
e) $f(x) = \ln|x - 2|$ for $x > 2$ and $f(x) = \ln|x - 2| + 1 - \ln 2$ for $x < 2$

a) For $x > 0$ , $|x| = x$ and $(\ln x)' = \frac{1}{x}$ from the table. For $x < 0$ , $|x| = -x > 0$ and, by the chain rule with inner function $u = -x$ , $u' = -1$ : $(\ln(-x))' = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$ . So $\ln|x|$ is an antiderivative of $\frac{1}{x}$ on $(-\infty, 0)$ AND on $(0, \infty)$ . Writing $\ln x$ alone covers only the right half: on $(-\infty, 0)$ , $\ln x$ is not even defined, while $\frac{1}{x}$ is.

b) The orange right branch passes through $(1, 1)$ and is the blue one shifted up by $1$ : $G(x) = \ln x + 1$ for $x > 0$ . The left branch passes through $(-1, -1)$ where $\ln|{-1}| = 0$ , so it is shifted DOWN by $1$ : $G(x) = \ln(-x) - 1$ for $x < 0$ . On each side, adding a constant does not change the derivative, so $G'(x) = \frac{1}{x}$ for every $x \ne 0$ . Yet $G(x) - \ln|x|$ equals $1$ for $x > 0$ and $-1$ for $x < 0$ : not constant. No contradiction, because the domain $x \ne 0$ is not an interval, and the Mean Value Theorem only works between two points of an interval on which the function is differentiable; $1$ and $-1$ are separated by the hole at $0$ . The most general antiderivative of $\frac{1}{x}$ is $\ln|x| + C_1$ for $x > 0$ and $\ln|x| + C_2$ for $x < 0$ , two independent constants.

c) On each interval, $f(x) = -\frac{1}{x} + C$ , since $\left(-x^{-1}\right)' = x^{-2}$ . For $x > 0$ : $f(1) = -1 + C_1 = 0$ , so $C_1 = 1$ . For $x < 0$ : $f(-1) = 1 + C_2 = 0$ , so $C_2 = -1$ . Hence $f(x) = 1 - \frac{1}{x}$ for $x > 0$ and $f(x) = -1 - \frac{1}{x}$ for $x < 0$ . Check: $f'(x) = \frac{1}{x^2}$ on both sides, $f(1) = 0$ , $f(-1) = -1 + 1 = 0$ . The classmate forced ONE constant on a domain with two pieces; with one condition per piece, the problem has exactly one solution. The two conditions are not redundant: $f(1) = 0$ says nothing at all about the left branch.

d) The function $\sec^2 x$ is undefined at $\frac{\pi}{2}$ , so the domain consists of the two intervals $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ , and $f(x) = \tan x + C$ on each, with its own constant. $f(0) = \tan 0 + C_1 = 0$ gives $C_1 = 0$ ; $f(\pi) = \tan\pi + C_2 = 0 + C_2 = 1$ gives $C_2 = 1$ . So $f = \tan x$ on the first interval and $\tan x + 1$ on the second. A single formula $\tan x + C$ would require $C = 0$ and $C = 1$ at once: the gap at $\frac{\pi}{2}$ , a vertical asymptote of $\tan$ , is what makes room for two constants.

e) By the linear rule, $(\ln|x - 2|)' = \frac{1}{x - 2}$ , with inner derivative $1$ . On $(2, \infty)$ : $f(3) = \ln 1 + C_1 = C_1 = 0$ . On $(-\infty, 2)$ : $f(0) = \ln|{-2}| + C_2 = \ln 2 + C_2 = 1$ , so $C_2 = 1 - \ln 2$ . Hence $f(x) = \ln(x - 2)$ for $x > 2$ and $f(x) = \ln(2 - x) + 1 - \ln 2$ for $x < 2$ . Check at $x = 0$ : $\ln 2 + 1 - \ln 2 = 1$ . The exact constant is $1 - \ln 2$ ; its size, about $0.31$ with $\ln 2 \approx 0.69$ , is a check, never the answer.

Exercise 6: Sketching an antiderivative from the graph of f: slopes, not guesses

The figure shows the graph of a continuous function $f$ on $[0, 6]$ , made of two line segments: from $(0, 2)$ to $(2, -2)$ , then from $(2, -2)$ to $(6, 2)$ . Let $F$ be the antiderivative of $f$ on $[0, 6]$ such that $F(0) = 0$ .

The graph of $f$ is the graph of the SLOPES of $F$ : the sign of $f$ gives the direction of $F$ , and the direction of $f$ gives the concavity of $F$ . Values of $F$ come from formulas, found piece by piece and joined by continuity.

a) On which intervals is $F$ increasing, decreasing? Where does $F$ have a local maximum, a local minimum? Justify with $f$ only.
b) Where is $F$ concave up, concave down? Give the inflection point of $F$ .
c) Read the two formulas of $f$ on the figure, and find $F$ on $[0, 2]$ , then on $[2, 6]$ . Explain how the second constant is found.
d) Compute $F(1)$ , $F(2)$ , $F(4)$ and $F(6)$ , and sketch the graph of $F$ .
e) A student's sketch shows a corner in the graph of $F$ at $x = 2$ , because the graph of $f$ has one there. Correct the sketch, and say what the corner of $f$ becomes on the graph of $F$ .

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Answers

a) $F$ increases on $[0, 1]$ and $[4, 6]$ , decreases on $[1, 4]$ ; local max at $x = 1$ , local min at $x = 4$
b) Concave down on $(0, 2)$ , concave up on $(2, 6)$ ; inflection point $(2, 0)$
c) $F(x) = 2x - x^2$ on $[0, 2]$ ; $F(x) = \frac{x^2}{2} - 4x + 6$ on $[2, 6]$ , constant from $F$ continuous at $2$
d) $F(1) = 1$ , $F(2) = 0$ , $F(4) = -2$ , $F(6) = 0$
e) No corner: $F'(2) = f(2) = -2$ from both sides; the corner of $f$ is the inflection point of $F$ .

a) $F' = f$ , so $F$ increases where $f > 0$ and decreases where $f < 0$ . The first segment crosses the axis at $x = 1$ , the second at $x = 4$ (from $-2$ at $x = 2$ to $2$ at $x = 6$ , it rises $1$ per unit). So $f > 0$ on $[0, 1)$ , $f < 0$ on $(1, 4)$ , $f > 0$ on $(4, 6]$ : $F$ increases on $[0, 1]$ , decreases on $[1, 4]$ , increases on $[4, 6]$ . By the first derivative test, $F$ has a local maximum at $x = 1$ , where $f$ changes from $+$ to $-$ , and a local minimum at $x = 4$ , where $f$ changes from $-$ to $+$ . Note what was NOT used: the height of $f$ . The highest point of $f$ , at $x = 0$ , is not an extremum of $F$ at all.

b) $F'' = f'$ : $F$ is concave up where $f$ increases and concave down where $f$ decreases. On $(0, 2)$ , $f$ decreases (slope $-2$ ), so $F$ is concave down; on $(2, 6)$ , $f$ increases (slope $1$ ), so $F$ is concave up. The concavity changes at $x = 2$ , where $f$ reaches its minimum, and $F$ is continuous there: $x = 2$ is the inflection point, and d) gives $F(2) = 0$ . The reading to memorize: the zeros of $f$ give the extrema of $F$ , the extrema of $f$ give the inflection points of $F$ .

c) The first segment passes through $(0, 2)$ with slope $-2$ : $f(x) = 2 - 2x$ on $[0, 2]$ . The second passes through $(4, 0)$ with slope $1$ : $f(x) = x - 4$ on $[2, 6]$ . On $[0, 2]$ , $F(x) = 2x - x^2 + C_1$ , and $F(0) = 0$ gives $C_1 = 0$ . On $[2, 6]$ , $F(x) = \frac{x^2}{2} - 4x + C_2$ . There is no condition given at a point of $[2, 6]$ , but $F$ is differentiable, hence CONTINUOUS, at $x = 2$ : the two formulas must agree there. $2 \cdot 2 - 4 = 0$ on the left, $2 - 8 + C_2$ on the right, so $C_2 = 6$ and $F(x) = \frac{x^2}{2} - 4x + 6$ on $[2, 6]$ . Continuity is the condition that fixes the constant of every piece after the first.

d) $F(1) = 2 - 1 = 1$ , $F(2) = 4 - 4 = 0$ , $F(4) = 8 - 16 + 6 = -2$ , $F(6) = 18 - 24 + 6 = 0$ . The sketch, shown in the solution figure: starting at the origin, $F$ rises to its local maximum $(1, 1)$ along an arc that bends down, falls through $(2, 0)$ , where the bending changes, to its local minimum $(4, -2)$ , then rises back to $(6, 0)$ bending up. Every feature of the sketch comes from a) and b); the four values of d) only place it. Check on one piece: $F'(x) = x - 4 = f(x)$ on $[2, 6]$ .

e) $F$ is an antiderivative of the continuous function $f$ , so $F$ is differentiable at $x = 2$ with $F'(2) = f(2) = -2$ . Both formulas agree: on the left, $F'(x) = 2 - 2x$ gives $-2$ ; on the right, $F'(x) = x - 4$ gives $-2$ . The graph of $F$ has ONE tangent line at $(2, 0)$ , of slope $-2$ : no corner. A corner in the graph of $F$ would need a JUMP in the values of $f$ . The corner of $f$ is a jump in the slope of $f$ , that is in $F''$ , which goes from $-2$ to $1$ : it shows up as the change of concavity, the inflection point of b). Each derivative smooths the picture by one level: a jump in $f$ gives a corner in $F$ , a corner in $f$ gives only a change of bending.

Exercise 7: A family of curves: which conditions pick one member, and which pick none

The antiderivatives of $f'(x) = 3x^2 - 3$ form the family $y = x^3 - 3x + C$ . The figure shows three members, for $C = 2$ , $C = 0$ and $C = -2$ , and the point $P(2, 1)$ .

Every member is a vertical translate of the others. One condition on $f$ picks one member; a second derivative needs two conditions, and those two conditions must actually reach the two constants.

a) Explain, with $f'$ , why all the members have the same slope at a given $x$ , and why no two of them ever meet. Which member passes through $P$ ?
b) Which member has its local minimum ON the $x$ -axis? For which values of $C$ does the member have three $x$ -intercepts?
c) Find $f$ if $f''(x) = 6x$ , $f(0) = 1$ and $f(1) = 0$ .
d) Same equation $f''(x) = 6x$ , with the conditions $f'(-1) = 3$ and $f'(1) = 3$ . How many solutions are there? Same question with $f'(-1) = 3$ and $f'(1) = 5$ .
e) Find $f$ if $f''(x) = 12x - 4$ and the graph of $f$ is tangent to the line $y = 5x - 3$ at the point where $x = 1$ .

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Answers

a) Slope $3x^2 - 3$ for every member; two members differ by a nonzero constant; through $P$ : $C = -1$
b) $C = 2$ ; three intercepts exactly when $-2 < C < 2$
c) $f(x) = x^3 - 2x + 1$
d) Infinitely many: $f(x) = x^3 + D$ ; none with $f'(1) = 5$
e) $f(x) = 2x^3 - 2x^2 + 3x - 1$

a) Every member has derivative $3x^2 - 3$ , so at a given $x$ all the members have the same slope: their tangent lines above that $x$ are parallel, which is why the curves look like copies slid up and down. Two members differ by the constant $C_1 - C_2 \ne 0$ at EVERY $x$ , so they never meet: exactly one member passes through each point of the plane. For $P(2, 1)$ : $8 - 6 + C = 1$ , so $C = -1$ , the member between the grey and green curves. The figure does not show it, and it does not need to: one point, one equation, one constant.

b) $F'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ vanishes at $x = \pm 1$ , and changes from $-$ to $+$ at $x = 1$ : the local minimum is $F(1) = 1 - 3 + C = C - 2$ , and the local maximum is $F(-1) = -1 + 3 + C = C + 2$ . The minimum is on the axis when $C = 2$ , the blue curve, which touches the axis at $(1, 0)$ . Three intercepts require the maximum above the axis and the minimum below: $C + 2 > 0$ and $C - 2 < 0$ , that is $-2 < C < 2$ . Indeed, the member then goes from $-\infty$ up to a positive maximum, down to a negative minimum and up to $+\infty$ , and the Intermediate Value Theorem gives one root on each of the three stretches. At $C = \pm 2$ one root is double; the grey member $C = 0$ has three: $x = 0$ and $x = \pm\sqrt 3$ .

c) $f'(x) = 3x^2 + C$ and $f(x) = x^3 + Cx + D$ . $f(0) = D = 1$ ; $f(1) = 1 + C + 1 = 0$ , so $C = -2$ . Hence $f(x) = x^3 - 2x + 1$ . Check: $f''(x) = 6x$ , $f(0) = 1$ , $f(1) = 1 - 2 + 1 = 0$ . Each condition was used for the constant it can reach: $f(0)$ kills $Cx$ and isolates $D$ , which is why it is used first.

d) $f'(x) = 3x^2 + C$ . Then $f'(-1) = 3 + C = 3$ and $f'(1) = 3 + C = 3$ give the SAME equation, $C = 0$ : the second condition repeats the first, because $f'$ is even. Then $f(x) = x^3 + D$ with $D$ free: infinitely many solutions, since no condition touches $D$ . With $f'(1) = 5$ instead, the conditions demand $C = 0$ and $C = 2$ at once: NO solution. Two conditions for two constants is necessary, not sufficient: both conditions on $f'$ leave $D$ untouched, and two conditions on the same quantity can repeat or contradict each other.

e) Tangent to $y = 5x - 3$ at $x = 1$ means two conditions: same slope, $f'(1) = 5$ , and same point, $f(1) = 5 - 3 = 2$ . Now $f'(x) = 6x^2 - 4x + C$ and $f'(1) = 2 + C = 5$ gives $C = 3$ . Then $f(x) = 2x^3 - 2x^2 + 3x + D$ and $f(1) = 3 + D = 2$ gives $D = -1$ . So $f(x) = 2x^3 - 2x^2 + 3x - 1$ . Check: $f''(x) = 12x - 4$ , $f'(1) = 6 - 4 + 3 = 5$ , $f(1) = 2 - 2 + 3 - 1 = 2$ . A tangency hidden in words is the most frequent way an exam gives two conditions in one sentence; reading only the point, and forgetting the slope, leaves $C$ undetermined.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 140 assignment, and each is false. Say what is wrong, settle it by differentiating or with a counterexample, and write the correct statement.

a) The most general antiderivative of $\frac{1}{x^3}$ is $\ln|x^3| + C$ .
b) An antiderivative of $\frac{1}{2x + 5}$ is $\ln|2x + 5|$ .
c) An antiderivative of $e^{x^2}$ is $\frac{e^{x^2}}{2x}$ , dividing by the derivative of the exponent as for $e^{2x}$ .
d) $\frac{(x + 1)^2}{2}$ and $\frac{x^2}{2} + x$ are different functions, so they cannot both be antiderivatives of $x + 1$ .
e) If $f$ is increasing on $\mathbb{R}$ , then every antiderivative of $f$ is increasing on $\mathbb{R}$ .

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Answers

a) False: $(\ln|x^3|)' = \frac{3}{x}$ . Correct: $-\frac{1}{2x^2} + C$ , one constant on each side of $0$ .
b) False: its derivative is $\frac{2}{2x + 5}$ . Correct: $\frac{1}{2}\ln|2x + 5|$ .
c) False: the derivative is $e^{x^2} - \frac{e^{x^2}}{2x^2}$ . Dividing by the inner derivative works for LINEAR inner functions only.
d) False: they differ by the constant $\frac{1}{2}$ , and both have derivative $x + 1$ .
e) False: $f(x) = x$ , $F(x) = \frac{x^2}{2}$ decreases on $(-\infty, 0]$ . $F$ increases where $f > 0$ .

a) FALSE. The exponent is $-3$ , not $-1$ : the power rule applies, $x^{-3} \mapsto \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$ . The logarithm is reserved for the exponent $-1$ alone. Differentiating the student's answer settles it: $\ln|x^3| = 3\ln|x|$ , whose derivative is $\frac{3}{x}$ , not $\frac{1}{x^3}$ ; at $x = 2$ , $\frac{3}{2}$ against $\frac{1}{8}$ . Correct statement: the most general antiderivative of $\frac{1}{x^3}$ is $-\frac{1}{2x^2} + C_1$ on $(0, \infty)$ and $-\frac{1}{2x^2} + C_2$ on $(-\infty, 0)$ . Check: $\left(-\frac{1}{2}x^{-2}\right)' = x^{-3}$ .

b) FALSE. By the chain rule with inner function $2x + 5$ , whose derivative is $2$ : $(\ln|2x + 5|)' = \frac{2}{2x + 5}$ , twice the target. Correct statement: $\frac{1}{2}\ln|2x + 5|$ is an antiderivative of $\frac{1}{2x + 5}$ on each of the intervals $x > -\frac{5}{2}$ and $x < -\frac{5}{2}$ , which is the linear rule $\frac{1}{a}F(ax + b)$ with $a = 2$ . The missing $\frac{1}{a}$ is the single most frequent error of the chapter, and differentiating catches it in one line every time.

c) FALSE. By the quotient rule, $\left(\frac{e^{x^2}}{2x}\right)' = \frac{2x e^{x^2} \cdot 2x - 2e^{x^2}}{4x^2} = e^{x^2} - \frac{e^{x^2}}{2x^2}$ : an extra term appears, because the $x$ in the denominator is NOT a constant and gets differentiated too. Dividing by the derivative of the inside works for $e^{2x}$ because that derivative is the constant $2$ , which passes through the differentiation untouched. Correct statement: for a LINEAR inside $ax + b$ , $\frac{1}{a}F(ax + b)$ is an antiderivative of $f(ax + b)$ ; for any other inside, the division fails. In fact $e^{x^2}$ , which is continuous, does have antiderivatives, but none can be written with the functions of this course.

d) FALSE. Expand: $\frac{(x + 1)^2}{2} = \frac{x^2}{2} + x + \frac{1}{2}$ . The two functions differ by the constant $\frac{1}{2}$ , and both have derivative $x + 1$ : both are antiderivatives, two members of the same family. Correct statement: two antiderivatives of the same function on an interval differ by a constant, and they may LOOK different. That is why two correct answers on a test can disagree in form: compare them by subtracting, or by differentiating both, never by their appearance.

e) FALSE. Take $f(x) = x$ , increasing on $\mathbb{R}$ . Its antiderivative $F(x) = \frac{x^2}{2}$ decreases on $(-\infty, 0]$ , where $f < 0$ . What decides whether $F$ increases is the SIGN of $f = F'$ , not its direction. Correct statement: an antiderivative of $f$ increases on any interval where $f > 0$ and decreases where $f < 0$ ; that $f$ increases says that $F'$ increases, that is, $F$ is concave up. This is the same reading as in Exercise 6: zeros and sign of $f$ for the direction of $F$ , slope of $f$ for its bending.

Exercise 9: A stone thrown up from a bridge: the equations of motion rebuilt from the acceleration

A stone is thrown straight up at $14.7$ m/s from a bridge whose parapet is $44.1$ m above a river, and falls into the water, missing the bridge. Air resistance is neglected, so the acceleration is $-9.8$ m/s $^2$ throughout the flight. Measure the height $s(t)$ , in metres, upward from the surface of the water, with $t$ in seconds from the throw.

No formula of kinematics is given: velocity and position are recovered from the acceleration by two antiderivatives, $v' = a$ and $s' = v$ , each with its constant. Every time asked for is exact.

a) Find $v(t)$ and $s(t)$ , and say what each constant of antidifferentiation represents.
b) When does the stone reach its highest point, and how high above the water is it then?
c) When does it pass back at the level of the parapet, and with what velocity?
d) When does it hit the water? Give the exact time and the exact velocity at impact, then estimate both with $\sqrt 5 \approx 2.236$ .
e) With what initial speed should the stone be thrown up, from the same parapet, to hit the water exactly $5$ s after the throw?

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Answers

a) $v(t) = 14.7 - 9.8t$ , $s(t) = 44.1 + 14.7t - 4.9t^2$ ; the constants are $v(0)$ and $s(0)$
b) At $t = 1.5$ s, at $55.125$ m above the water
c) At $t = 3$ s, with $v(3) = -14.7$ m/s
d) $t = \frac{3 + 3\sqrt 5}{2} \approx 4.85$ s, $v = -14.7\sqrt 5 \approx -32.9$ m/s
e) $v_0 = 15.68$ m/s

a) $v'(t) = a(t) = -9.8$ , so $v(t) = -9.8t + C_1$ . At $t = 0$ , $v(0) = C_1$ : the first constant IS the initial velocity, $C_1 = 14.7$ , positive because the throw is upward on an upward axis. Then $s'(t) = v(t) = 14.7 - 9.8t$ gives $s(t) = 14.7t - 4.9t^2 + C_2$ , and $s(0) = C_2$ is the initial height, $44.1$ . So $v(t) = 14.7 - 9.8t$ and $s(t) = 44.1 + 14.7t - 4.9t^2$ . Check: $s'' = -9.8$ , $s(0) = 44.1$ , $s'(0) = 14.7$ . The formula $s = s_0 + v_0 t + \frac{1}{2}at^2$ of physics is exactly this computation with letters, and it holds only while $a$ is constant, which Exercise 10 will not be.

b) At the highest point the stone stops rising: $v(t) = 0$ , so $t = \frac{14.7}{9.8} = 1.5$ s. Then $s(1.5) = 44.1 + 14.7 \cdot 1.5 - 4.9 \cdot 2.25 = 44.1 + 22.05 - 11.025 = 55.125$ m above the water, that is $11.025$ m above the parapet. It is a maximum because $v > 0$ before $t = 1.5$ and $v < 0$ after: the first derivative test, with $s' = v$ . The trap is to give $11.025$ m, the rise above the hand, when the question measures from the water: the axis was fixed in the statement, and $s$ is measured on it.

c) $s(t) = 44.1$ means $14.7t - 4.9t^2 = 0$ , that is $4.9t(3 - t) = 0$ : $t = 0$ , the throw, or $t = 3$ s. Then $v(3) = 14.7 - 29.4 = -14.7$ m/s: the same speed as the throw, downward. The flight above the parapet is symmetric about the top, $1.5$ s up and $1.5$ s down, because $s$ is a parabola whose axis is $t = 1.5$ . The sign of $v$ is part of the answer: $-14.7$ m/s says down, and a velocity given as $14.7$ m/s is a speed, which is a different word.

d) $s(t) = 0$ : $4.9t^2 - 14.7t - 44.1 = 0$ . Dividing by $4.9$ , since $14.7 = 3 \cdot 4.9$ and $44.1 = 9 \cdot 4.9$ : $t^2 - 3t - 9 = 0$ , so $t = \frac{3 \pm \sqrt{9 + 36}}{2} = \frac{3 \pm 3\sqrt 5}{2}$ . The negative root is rejected, the flight starts at $t = 0$ : $t = \frac{3 + 3\sqrt 5}{2}$ s. Then $v = 14.7 - 9.8 \cdot \frac{3 + 3\sqrt 5}{2} = 14.7 - 14.7 - 14.7\sqrt 5 = -14.7\sqrt 5$ m/s. Estimates: $t \approx \frac{3 + 6.708}{2} \approx 4.85$ s and $v \approx -32.9$ m/s. Check by energy, independent of the time: $v^2 = v_0^2 + 2 \cdot 9.8 \cdot 44.1 = 216.09 + 864.36 = 1080.45 = 5 \cdot 14.7^2$ . Dividing by $4.9$ first is what makes the computation possible by hand.

e) Keep the initial speed as the unknown $v_0$ : the same two antiderivatives give $s(t) = 44.1 + v_0 t - 4.9t^2$ . The condition $s(5) = 0$ reads $44.1 + 5v_0 - 122.5 = 0$ , so $5v_0 = 78.4$ and $v_0 = 15.68$ m/s. Check: $s(5) = 44.1 + 78.4 - 122.5 = 0$ . This is an initial value problem run backwards: the condition at $t = 5$ fixes the constant $C_1 = v_0$ instead of the value at $t = 0$ . Faster than the $14.7$ m/s of the statement, as expected, since the flight is longer than $4.85$ s.

Exercise 10: A final exam question: stopping distance when the braking builds up

A car travels at $18$ m/s, that is $64.8$ km/h, on a straight road. At $t = 0$ the driver brakes. The deceleration is not instantaneous: it grows linearly during the first second, $a(t) = -6t$ m/s $^2$ for $0 \le t \le 1$ , then stays at $a(t) = -6$ m/s $^2$ until the car stops, as the figure shows.

Measure the position $s(t)$ , in metres, from the point where braking starts, in the direction of motion. This is the shape of a long final exam question: two pieces, two antiderivatives on each, constants fixed by continuity, and no calculator.

a) Find $v(t)$ and $s(t)$ for $0 \le t \le 1$ , then $v(1)$ and $s(1)$ .
b) Find $v(t)$ and $s(t)$ for $t \ge 1$ , until the car stops. Explain how the constants are found, and give the time at which the car stops.
c) Find the stopping distance.
d) Compare with an ideal braking at $-6$ m/s $^2$ from $t = 0$ . How many metres does the ramp cost, and why?
e) With the same braking profile, what is the largest initial speed $v_0$ for which the stopping distance is at most $20$ m? Give the exact value, then bracket it without a calculator.

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Answers

a) $v(t) = 18 - 3t^2$ , $s(t) = 18t - t^3$ ; $v(1) = 15$ m/s, $s(1) = 17$ m
b) $v(t) = 21 - 6t$ , $s(t) = -3t^2 + 21t - 1$ by continuity at $t = 1$ ; stop at $t = 3.5$ s
c) $s(3.5) = 35.75$ m
d) $27$ m for the ideal braking: the ramp costs $8.75$ m
e) $v_0 = -3 + 6\sqrt 7$ m/s, between $12.84$ and $12.9$ m/s, about $46$ km/h

a) On $[0, 1]$ , $v'(t) = -6t$ gives $v(t) = -3t^2 + C_1$ , and $v(0) = 18$ gives $C_1 = 18$ : $v(t) = 18 - 3t^2$ . Then $s'(t) = v(t)$ gives $s(t) = 18t - t^3 + C_2$ , and $s(0) = 0$ gives $C_2 = 0$ . So $v(1) = 15$ m/s and $s(1) = 18 - 1 = 17$ m. During the ramp the car loses only $3$ m/s in the first second, half of what a full braking would take away: the acceleration is not constant, so no equation of constant acceleration may be used on this piece, only antiderivatives.

b) On $[1, \infty)$ , $v'(t) = -6$ gives $v(t) = -6t + C_3$ . No value of $v$ is given at a time of this piece, but the velocity is continuous (it cannot jump without an infinite force): $v(1) = 15$ , so $-6 + C_3 = 15$ and $C_3 = 21$ , $v(t) = 21 - 6t$ . Then $s(t) = 21t - 3t^2 + C_4$ , and continuity of the position at $t = 1$ gives $21 - 3 + C_4 = 17$ , so $C_4 = -1$ : $s(t) = -3t^2 + 21t - 1$ . The car stops when $v(t) = 0$ : $t = 3.5$ s. Four constants, four conditions: two initial values and two junction conditions, exactly as in Exercise 6.

c) The car moves forward during the whole braking ( $v > 0$ on $[0, 3.5)$ ), so the stopping distance is $s(3.5) = -3 \cdot 12.25 + 73.5 - 1 = -36.75 + 72.5 = 35.75$ m. Check: $s'(t) = 21 - 6t = v(t)$ and $s(1) = -3 + 21 - 1 = 17$ , the value of a). Using the second formula before $t = 1$ , or the first after, is the classic error of piecewise motion: each formula is valid on its own piece only.

d) Ideal braking: $v(t) = 18 - 6t$ and $s(t) = 18t - 3t^2$ , from the same two antiderivatives. The car stops at $t = 3$ s, after $s(3) = 54 - 27 = 27$ m. The ramp costs $35.75 - 27 = 8.75$ m, about two car lengths. Why: after one second the ramped car still moves at $15$ m/s, having covered $17$ m, while the ideal car is at $12$ m/s after $15$ m. From there both decelerate at $6$ m/s $^2$ ; a car at speed $u$ then needs $\frac{u^2}{12}$ metres, since $v = u - 6\tau$ stops at $\tau = \frac{u}{6}$ after $u\tau - 3\tau^2 = \frac{u^2}{12}$ . That is $18.75$ m for the ramped car against $12$ m. The whole loss is decided in the first second.

e) For an initial speed $v_0 > 3$ , the ramp gives $v(t) = v_0 - 3t^2$ and $s(t) = v_0 t - t^3$ , so $v(1) = v_0 - 3$ and $s(1) = v_0 - 1$ ; the remaining distance is $\frac{(v_0 - 3)^2}{12}$ by d). The stopping distance is $D(v_0) = v_0 - 1 + \frac{(v_0 - 3)^2}{12}$ , increasing in $v_0$ . Check: $D(18) = 17 + \frac{225}{12} = 35.75$ . Solve $D(v_0) = 20$ : $12v_0 - 12 + v_0^2 - 6v_0 + 9 = 240$ , that is $v_0^2 + 6v_0 - 243 = 0$ , so $v_0 = -3 + \sqrt{9 + 243} = -3 + \sqrt{252} = -3 + 6\sqrt 7$ m/s, the negative root rejected. Bracket: $2.64^2 = 6.9696 < 7 < 7.0225 = 2.65^2$ , so $15.84 < 6\sqrt 7 < 15.9$ and $12.84 < v_0 < 12.9$ m/s, about $46$ km/h. It exceeds $3$ , as the formula for $D$ assumed. The fastest speed for a $20$ m stop is far below the $18$ m/s of the statement.

Corrected exercises: antiderivatives (MATH 140)

Course recap

Part A: the basics (/50)

Exercise 1: The general antiderivative: read the table backwards, then differentiate

Exercise 2: Rewrite before antidifferentiating: no product rule and no quotient rule backwards

Exercise 3: Trigonometric functions: an identity first, then the table

Exercise 4: Initial value problems: one condition for each constant

Exercise 5: One constant per interval: when the domain has a gap

Part B: problems and reasoning (/50)

Exercise 6: Sketching an antiderivative from the graph of f: slopes, not guesses

Exercise 7: A family of curves: which conditions pick one member, and which pick none

Exercise 8: Five statements to correct

Exercise 9: A stone thrown up from a bridge: the equations of motion rebuilt from the acceleration

Exercise 10: A final exam question: stopping distance when the braking builds up

See also

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