Corrected exercises: antiderivatives (MATH 140)

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.

$5 \cdot \frac{1}{x}$ 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.

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.

$(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$ 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 $b^x$.

$F_1'(x) = \left(1 \cdot \ln x + x \cdot \frac{1}{x}\right) - 1 + 0 = \ln x$, 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.

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.

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.

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.

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 $1$.

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.

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.

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.

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.

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 $\cos^2 x$.

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.

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.

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.

$f(x) = \arctan x + C$ 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 $\frac{\pi}{4}$.

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.

$f'(x) = 4x^3 + 2e^{2x} + C_1$, 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.