Chapter 12 — Exercises

38 problems on antiderivatives, indefinite integrals, initial value problems, and reconstruction from rates. Difficulty runs ⭐ (recall the table) to ⭐⭐⭐⭐ (multi-step modeling).

Every antiderivative can be checked by differentiating it back to the integrand (Section 12.13). The solutions do this for you on the harder problems; do it yourself on the easy ones.

Tier Count Problems
⭐ 9 A1–A6, B1, C1, D1
⭐⭐ 14 A7–A9, B2–B4, C2–C3, D2–D3, E1–E2, F1–F2
⭐⭐⭐ 11 B5, C4, E3–E4, F3–F4, G1–G3, H1–H2
⭐⭐⭐⭐ 4 E5, G4, H3, H4
Total 38

Part A — The Basic Table and the Power Rule (Sections 12.4–12.6)

A1 ⭐ $\displaystyle\int 7\,dx$.

Solution$7x + C$. (A constant integrates to a linear term; check $(7x)' = 7$.)

A2 ⭐ $\displaystyle\int x^5\,dx$.

SolutionPower rule: raise the exponent, divide by the new exponent. $\dfrac{x^6}{6} + C$.

A3 ⭐ $\displaystyle\int (4x - 3)\,dx$.

Solution$2x^2 - 3x + C$. Check: $(2x^2 - 3x)' = 4x - 3$. ✓

A4 ⭐ $\displaystyle\int (6x^2 + 2x + 1)\,dx$.

Solution$2x^3 + x^2 + x + C$.

A5 ⭐ $\displaystyle\int \frac{1}{x}\,dx$.

SolutionThe $n=-1$ exception (Section 12.6): $\ln\lvert x\rvert + C$. Not a power of $x$.

A6 ⭐ $\displaystyle\int x^{-3}\,dx$.

SolutionHere $n=-3 \neq -1$, so the power rule applies: $\dfrac{x^{-2}}{-2} + C = -\dfrac{1}{2x^2} + C$.

A7 ⭐⭐ $\displaystyle\int \sqrt[3]{x}\,dx$.

SolutionWrite as $x^{1/3}$: $\dfrac{x^{4/3}}{4/3} + C = \dfrac{3}{4}x^{4/3} + C$. (Dividing by $4/3$ means multiplying by $3/4$.)

A8 ⭐⭐ $\displaystyle\int \left(\sqrt{x} + \frac{1}{x^2}\right)\,dx$.

Solution$\int x^{1/2}\,dx + \int x^{-2}\,dx = \dfrac{2}{3}x^{3/2} - \dfrac{1}{x} + C$.

A9 ⭐⭐ $\displaystyle\int \frac{2x^3 - x + 5}{x}\,dx$.

SolutionDivide first (no quotient rule for integrals!): $\dfrac{2x^3 - x + 5}{x} = 2x^2 - 1 + \dfrac{5}{x}$. So $\int = \dfrac{2}{3}x^3 - x + 5\ln\lvert x\rvert + C$.

Part B — Trigonometric and Inverse-Trig Antiderivatives (Section 12.4)

B1 ⭐ $\displaystyle\int \sin x\,dx$.

Solution$-\cos x + C$. The minus sign lives with $\sin$. Check: $(-\cos x)' = \sin x$. ✓

B2 ⭐⭐ $\displaystyle\int (3\cos x - 2\sin x)\,dx$.

Solution$3\sin x + 2\cos x + C$. (The $-2\sin x$ integrates to $-2(-\cos x) = +2\cos x$.)

B3 ⭐⭐ $\displaystyle\int \sec^2 x\,dx$.

Solution$\tan x + C$, straight from the table.

B4 ⭐⭐ $\displaystyle\int (\sec x \tan x + \sec^2 x)\,dx$.

Solution$\sec x + \tan x + C$.

B5 ⭐⭐⭐ $\displaystyle\int \frac{3}{1+x^2}\,dx + \int \frac{1}{\sqrt{1-x^2}}\,dx$.

Solution$3\arctan x + \arcsin x + C$. (Two table entries plus linearity; one combined constant.)

Part C — Exponentials and Logarithms (Sections 12.4, 12.6)

C1 ⭐ $\displaystyle\int e^x\,dx$.

Solution$e^x + C$. The function that is its own antiderivative.

C2 ⭐⭐ $\displaystyle\int \left(2e^x + \frac{5}{x}\right)\,dx$.

Solution$2e^x + 5\ln\lvert x\rvert + C$.

C3 ⭐⭐ $\displaystyle\int 3^x\,dx$.

Solution$\dfrac{3^x}{\ln 3} + C$. Check: $\left(\dfrac{3^x}{\ln 3}\right)' = \dfrac{3^x \ln 3}{\ln 3} = 3^x$. ✓

C4 ⭐⭐⭐ $\displaystyle\int \left(e^x - \frac{1}{x} + x^{-1/2}\right)\,dx$.

Solution$e^x - \ln\lvert x\rvert + 2x^{1/2} + C$. The middle term is the $n=-1$ log; the last is the power rule with $n=-\tfrac12$.

Part D — Linearity and Simplify-First (Section 12.7)

D1 ⭐ $\displaystyle\int (x^2 + 1)^2\,dx$.

SolutionExpand first (there is no product rule for integrals): $(x^2+1)^2 = x^4 + 2x^2 + 1$. So $\int = \dfrac{x^5}{5} + \dfrac{2x^3}{3} + x + C$.

D2 ⭐⭐ $\displaystyle\int x^2(3x - 4)\,dx$.

SolutionDistribute: $3x^3 - 4x^2$. Then $\int = \dfrac{3x^4}{4} - \dfrac{4x^3}{3} + C$.

D3 ⭐⭐ Explain in one sentence why $\displaystyle\int x \cdot e^x\,dx \neq \left(\int x\,dx\right)\left(\int e^x\,dx\right)$.

SolutionLinearity covers sums and constant multiples only — there is **no product rule for integrals** (Section 12.7). Test: the right side is $\tfrac{x^2}{2}e^x$, whose derivative (by the product rule) is $x e^x + \tfrac{x^2}{2}e^x \neq x e^x$. (The correct antiderivative, $(x-1)e^x + C$, needs integration by parts from Chapter 15.)

Part E — Reading the Chain Rule Backward (Section 12.7½)

These integrands are standard functions of a linear inner function $ax+b$. Integrate as usual, then divide by the inner slope $a$. The trick works only for linear interiors.

E1 ⭐⭐ $\displaystyle\int e^{2x}\,dx$.

Solution$\dfrac{1}{2}e^{2x} + C$. Check: $\left(\tfrac12 e^{2x}\right)' = \tfrac12 \cdot 2 e^{2x} = e^{2x}$. ✓

E2 ⭐⭐ $\displaystyle\int \cos(5x)\,dx$.

Solution$\dfrac{1}{5}\sin(5x) + C$. Divide by the inner slope $5$ to cancel the chain factor.

E3 ⭐⭐⭐ $\displaystyle\int (3x + 1)^4\,dx$.

SolutionPower rule then divide by the slope $3$: $\dfrac{(3x+1)^5}{3 \cdot 5} + C = \dfrac{(3x+1)^5}{15} + C$. Check: derivative is $\tfrac{5(3x+1)^4 \cdot 3}{15} = (3x+1)^4$. ✓

E4 ⭐⭐⭐ $\displaystyle\int \frac{1}{2x - 7}\,dx$.

SolutionThe $n=-1$ case with linear interior: $\dfrac{1}{2}\ln\lvert 2x - 7\rvert + C$.

E5 ⭐⭐⭐⭐ $\displaystyle\int \frac{1}{9 + x^2}\,dx$. Then explain why the same method fails for $\displaystyle\int \cos(x^2)\,dx$.

SolutionRescaled arctangent with $a^2 = 9$, so $a = 3$: $\dfrac{1}{3}\arctan\dfrac{x}{3} + C$. Check: $\dfrac{d}{dx}\left[\tfrac13\arctan(x/3)\right] = \tfrac13 \cdot \dfrac{1/3}{1+(x/3)^2} = \dfrac{1}{9+x^2}$. ✓

For $\cos(x^2)$ the inner function $x^2$ is **not linear**, so its slope $2x$ is not a constant you can divide out. The divide-by-the-slope trick applies only to $ax+b$. In fact $\int \cos(x^2)\,dx$ is non-elementary — a Fresnel integral (Section 12.9).

Part F — Initial Value Problems (Section 12.8)

F1 ⭐⭐ Find $f$ if $f'(x) = 3x^2$ and $f(0) = 5$.

Solution$f(x) = x^3 + C$; $f(0) = C = 5$, so $f(x) = x^3 + 5$.

F2 ⭐⭐ Find $f$ if $f'(x) = \cos x$ and $f(\pi/2) = 0$.

Solution$f(x) = \sin x + C$; $f(\pi/2) = 1 + C = 0 \Rightarrow C = -1$. So $f(x) = \sin x - 1$.

F3 ⭐⭐⭐ Find $f$ if $f''(x) = 6x$, $f'(0) = 2$, and $f(0) = 1$.

SolutionIntegrate once: $f'(x) = 3x^2 + C_1$; $f'(0) = C_1 = 2$, so $f'(x) = 3x^2 + 2$. Integrate again: $f(x) = x^3 + 2x + C_2$; $f(0) = C_2 = 1$. So $f(x) = x^3 + 2x + 1$. (Two integrations, two conditions — the bookkeeping balances.)

F4 ⭐⭐⭐ Find $f$ if $f'(x) = e^x + \dfrac{1}{x}$ and $f(1) = 2e$.

Solution$f(x) = e^x + \ln\lvert x\rvert + C$. Apply $f(1) = e + \ln 1 + C = e + C = 2e \Rightarrow C = e$. So $f(x) = e^x + \ln\lvert x\rvert + e$.

Part G — Rectilinear Motion (Section 12.10)

G1 ⭐⭐⭐ A particle has acceleration $a(t) = 6t$ m/s², with $v(0) = 1$ m/s and $s(0) = 4$ m. Find $v(t)$ and $s(t)$.

Solution$v(t) = 3t^2 + C_1$; $v(0) = C_1 = 1$, so $v(t) = 3t^2 + 1$. $s(t) = t^3 + t + C_2$; $s(0) = C_2 = 4$, so $s(t) = t^3 + t + 4$.

G2 ⭐⭐⭐ A car travels at $30$ m/s and brakes with constant deceleration $5$ m/s². Take $s(0)=0$. Find the stopping time and stopping distance.

Solution$a = -5$, so $v(t) = -5t + 30$ (using $v(0)=30$). Stops when $v=0$: $t = 6$ s. Position: $s(t) = -2.5t^2 + 30t$. Stopping distance $s(6) = -2.5(36) + 180 = -90 + 180 = 90$ m.

G3 ⭐⭐⭐ A ball is thrown straight up at $v_0 = 25$ m/s from a $50$ m cliff ($g = 9.8$ m/s²). Find $v(t)$ and $s(t)$, and when it hits the ground.

Solution$a = -9.8$, $v(t) = -9.8t + 25$, $s(t) = -4.9t^2 + 25t + 50$. Ground: $-4.9t^2 + 25t + 50 = 0 \Rightarrow 4.9t^2 - 25t - 50 = 0$. $t = \dfrac{25 + \sqrt{625 + 980}}{9.8} = \dfrac{25 + \sqrt{1605}}{9.8} \approx \dfrac{25 + 40.06}{9.8} \approx 6.64$ s.

G4 ⭐⭐⭐⭐ A car accelerates from rest at $3$ m/s² for $5$ s, then holds constant velocity. Find its position at $t = 10$ s. (Hint: the motion has two phases.)

Solution**Phase 1** ($0 \le t \le 5$): $v(t) = 3t$, $s(t) = 1.5t^2$. At $t=5$: $v = 15$ m/s, $s = 37.5$ m. **Phase 2** ($5 \le t \le 10$): constant $v = 15$, so position increases by $15 \times 5 = 75$ m. Total: $37.5 + 75 = 112.5$ m. (You must antidifferentiate each phase separately and stitch them with the end-of-phase-1 values as the new initial conditions.)

Part H — Reconstructing Totals from Rates, Across Fields (Section 12.11)

H1 ⭐⭐⭐ (Economics.) A firm's marginal cost is $C'(q) = 100 - 0.1q$ dollars per unit, with fixed cost $C(0) = 1000$. Find the total cost function and the cost of producing $200$ units.

Solution$C(q) = 100q - 0.05q^2 + 1000$ (the constant is the fixed cost). At $q=200$: $C(200) = 20000 - 0.05(40000) + 1000 = 20000 - 2000 + 1000 = 19000$ dollars.

H2 ⭐⭐⭐ (Physics / electricity.) Current into a capacitor is $I(t) = 4t$ amperes, and the initial charge is $Q(0) = 2$ coulombs. Since $Q'(t) = I(t)$, find the charge $Q(t)$.

Solution$Q(t) = \int 4t\,dt + Q_0 = 2t^2 + C$; $Q(0) = C = 2$, so $Q(t) = 2t^2 + 2$ coulombs. (Charge is the antiderivative of current — Section 12.11.)

H3 ⭐⭐⭐⭐ (Biology / pharmacokinetics.) A drug is infused so that the amount in the bloodstream changes at net rate $M'(t) = 50 - 4t$ mg/min for $0 \le t \le 10$, starting from $M(0) = 0$. Find $M(t)$, the time the bloodstream amount is maximized, and that maximum amount.

Solution$M(t) = 50t - 2t^2 + C$; $M(0) = C = 0$, so $M(t) = 50t - 2t^2$ mg. The amount peaks when the net rate is zero: $M'(t) = 50 - 4t = 0 \Rightarrow t = 12.5$ min — but that lies **outside** the interval $[0,10]$. On $[0,10]$, $M'(t) = 50 - 4t > 0$ throughout (it equals $10$ at $t=10$), so $M$ is still increasing at the end; the maximum on the interval is at $t = 10$: $M(10) = 500 - 200 = 300$ mg. (Lesson: always check whether the critical time lies in the valid domain.)

H4 ⭐⭐⭐⭐ (Conceptual / differential equations.) A reservoir loses water at a rate proportional to its current volume: $V'(t) = -0.01\,V(t)$, with $V(0) = 1000$ gal. Explain why this is not a plain antidifferentiation problem, and state its solution.

SolutionThe right-hand side contains the unknown $V$ itself, not a known function of $t$ alone, so you cannot simply integrate $-0.01V$ with respect to $t$ (Section 12.11 pitfall). This is a **differential equation** of the form $y' = ky$, solved by separation of variables in Chapter 19. Its solution is exponential decay: $V(t) = 1000\,e^{-0.01t}$ gal. (Check: $V'(t) = 1000(-0.01)e^{-0.01t} = -0.01\,V(t)$. ✓) Plain antidifferentiation solves only $y' = g(x)$ where the right side is a known function of $x$.

Reflection

Antidifferentiation is differentiation run backward, well-defined up to the constant $C$ that an initial condition pins down (Section 12.2, 12.8). You can now integrate every function in the basic table, dismantle sums with linearity, handle linear interiors by reading the chain rule backward, solve initial value problems, and reconstruct position from acceleration or any total from its rate.

The deliberately suggestive notation $\int$ — the elongated "S" for sum — is a promissory note. Chapter 13 cashes it in by defining the definite integral as the limit of a sum (area under a curve), and Chapter 14 delivers the Fundamental Theorem of Calculus, revealing that the antiderivatives you found here are exactly the tool for computing those areas (Section 12.15).