Calc 2 Full Practice Test

1. Partial Fractions — Setup Only Show first step only

1(a)

$\displaystyle\int \frac{3x^2 - x + 5}{(x-2)(x-3)^3}\,dx$ — show the partial fraction decomposition setup.

$\dfrac{A}{x-2} + \dfrac{B}{x-3} + \dfrac{C}{(x-3)^2} + \dfrac{D}{(x-3)^3}$

1(b)

$\displaystyle\int \frac{x^3 + 2x + 1}{(x-4)(x^2+9)^2}\,dx$ — show the partial fraction decomposition setup.

$\dfrac{A}{x-4} + \dfrac{Bx+C}{x^2+9} + \dfrac{Dx+E}{(x^2+9)^2}$

2. Trapezoidal Rule Numerical

2

Use the Trapezoidal Method with $n=4$ to approximate $\displaystyle\int_{-1}^{3} 3x^2\,dx$.

Recall: $T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f(x_n)]$

$\Delta x = 1$, nodes: $x = -1, 0, 1, 2, 3$
$f$ values: $3, 0, 3, 12, 27$
$T_4 = \frac{1}{2}[3 + 2(0) + 2(3) + 2(12) + 27] = \frac{1}{2}(60) = 30$

3. Simpson's Rule Numerical

3

Use Simpson's Method with $n=4$ to approximate $\displaystyle\int_{1}^{9} x^3\,dx$.

Recall: $S_n = \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)]$

$\Delta x = 2$, nodes: $1, 3, 5, 7, 9$
$f$ values: $1, 27, 125, 343, 729$
$S_4 = \frac{2}{3}[1 + 4(27) + 2(125) + 4(343) + 729]$
$= \frac{2}{3}[1 + 108 + 250 + 1372 + 729] = \frac{2}{3}(2460) = 1640$

4. Improper Integrals — Convergence/Divergence Comparison Test

4(a)

$\displaystyle\int_{5}^{\infty} \frac{x^{3/2}}{x^2+1}\,dx$ — converge or diverge?

For large $x$: $\frac{x^{3/2}}{x^2+1} \approx \frac{x^{3/2}}{x^2} = x^{-1/2}$.
$\int_5^\infty x^{-1/2}\,dx$ diverges (p-integral, $p = 1/2 \le 1$). By comparison, diverges.

4(b)

$\displaystyle\int_{2}^{\infty} \frac{1}{xe^x}\,dx$ — converge or diverge?

$\frac{1}{xe^x} \le \frac{1}{e^x} = e^{-x}$ for $x \ge 2$.
$\int_2^\infty e^{-x}\,dx$ converges. By comparison, converges.

4(c)

$\displaystyle\int_{3}^{\infty} \frac{1}{(\ln x)^5}\,dx$ — converge or diverge?

For $x \ge 3$: $\ln x \le x$, so $(\ln x)^5 \le x^5$, meaning $\frac{1}{(\ln x)^5} \ge \frac{1}{x^5}$... but that's the wrong direction.
Better: $\ln x \le x^{1/2}$ for large $x$, so $\frac{1}{(\ln x)^5} \ge \frac{1}{x^{5/2}}$. Hmm — actually use that $\ln x \lt x$, so $\frac{1}{(\ln x)^5} \gt \frac{1}{x^5}$ which converges — not useful.
Key insight: $\ln x$ grows slower than any power of $x$, so $\frac{1}{(\ln x)^5} \gg \frac{1}{x}$ for large $x$. Compare to $\frac{1}{x}$: since $\frac{1/(\ln x)^5}{1/x} = \frac{x}{(\ln x)^5} \to \infty$, the integrand is eventually larger than $\frac{1}{x}$. By limit comparison, diverges.

4(d)

$\displaystyle\int_{10}^{\infty} \frac{4x^2+3x}{x^{5/2}+x^{7/2}-2}\,dx$ — converge or diverge?

Dominant terms: numerator $\sim 4x^2$, denominator $\sim x^{7/2}$.
Ratio $\sim \frac{4x^2}{x^{7/2}} = 4x^{-3/2}$. $\int_{10}^\infty x^{-3/2}\,dx$ converges ($p = 3/2 > 1$). By limit comparison, converges.

5. Evaluate the Integral All methods

5(a) — Trig Sub

$\displaystyle\int \frac{1}{(x^2+9)^{3/2}}\,dx$

Let $x = 3\tan\theta$, $dx = 3\sec^2\theta\,d\theta$, $(x^2+9)^{3/2} = 27\sec^3\theta$.
$\displaystyle\int \frac{3\sec^2\theta}{27\sec^3\theta}\,d\theta = \frac{1}{9}\int\cos\theta\,d\theta = \frac{1}{9}\sin\theta + C$
Back-sub: $\sin\theta = \frac{x}{\sqrt{x^2+9}}$.
$\boxed{\dfrac{x}{9\sqrt{x^2+9}} + C}$

5(b) — Trig Sub

$\displaystyle\int \frac{\sqrt{x^2-16}}{x}\,dx,\quad x > 4$

Let $x = 4\sec\theta$, $\sqrt{x^2-16} = 4\tan\theta$.
$\displaystyle\int \frac{4\tan\theta}{4\sec\theta}\cdot 4\sec\theta\tan\theta\,d\theta = 4\int\tan^2\theta\,d\theta = 4(\tan\theta - \theta) + C$
$\boxed{\sqrt{x^2-16} - 4\sec^{-1}\!\left(\frac{x}{4}\right) + C}$

5(c) — Partial Fractions

$\displaystyle\int \frac{1}{x^3+x^2}\,dx$

Factor: $x^2(x+1)$. Decompose: $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$.
Solving: $A = -1,\ B = 1,\ C = 1$.
$\boxed{-\ln|x| - \frac{1}{x} + \ln|x+1| + C}$

5(d) — Polynomial Long Division + PF

$\displaystyle\int \frac{x^2+3x+2}{x^2+x+1}\,dx$

Degree equal — divide first: $\frac{x^2+3x+2}{x^2+x+1} = 1 + \frac{2x+1}{x^2+x+1}$.
Note $\frac{d}{dx}(x^2+x+1) = 2x+1$, so second term integrates to $\ln|x^2+x+1|$.
$\boxed{x + \ln|x^2+x+1| + C}$

5(e) — Partial Fractions

$\displaystyle\int \frac{8x+7}{x^2-8x+16}\,dx$

Denominator: $(x-4)^2$. Decompose: $\frac{A}{x-4}+\frac{B}{(x-4)^2}$.
$8x+7 = A(x-4)+B$: $A=8,\ B=39$.
$\boxed{8\ln|x-4| - \frac{39}{x-4} + C}$

5(f) — Substitution

$\displaystyle\int \frac{1}{1+\sqrt[3]{x}}\,dx$

Let $u = x^{1/3}$, so $x = u^3$, $dx = 3u^2\,du$.
$\displaystyle\int \frac{3u^2}{1+u}\,du$. Divide: $\frac{u^2}{1+u} = u - 1 + \frac{1}{1+u}$.
$\boxed{3\!\left(\frac{u^2}{2} - u + \ln|1+u|\right)+C}$, back-sub $u = x^{1/3}$.

5(g) — Improper + u-sub

$\displaystyle\int_{5}^{\infty} \frac{1}{x(\ln x)^3}\,dx$

Let $u = \ln x$, $du = dx/x$.
$\displaystyle\int_{\ln 5}^{\infty} u^{-3}\,du = \left[-\frac{1}{2u^2}\right]_{\ln 5}^{\infty} = \frac{1}{2(\ln 5)^2}$
Converges to $\boxed{\dfrac{1}{2(\ln 5)^2}}$

5(h) — Improper, discontinuity

$\displaystyle\int_{1}^{5} \frac{1}{(x-4)^2}\,dx$

Discontinuity at $x=4$ (interior). Split: $\int_1^4 + \int_4^5$.
$\int_1^4 \frac{dx}{(x-4)^2} = \left[-\frac{1}{x-4}\right]_1^4 \to \infty$.
Diverges.

6. Differential Equations Sep. of Variables

6(a)

$\dfrac{dy}{dx} = e^{x+y},\quad y(0) = 5$

Separate: $e^{-y}\,dy = e^x\,dx$.
Integrate: $-e^{-y} = e^x + C$.
IC: $-e^{-5} = 1 + C \Rightarrow C = -1 - e^{-5}$.
$\boxed{-e^{-y} = e^x - 1 - e^{-5}}$

6(b)

$\dfrac{dy}{dx} = \cos x\tan y,\quad y\!\left(\dfrac{3\pi}{2}\right) = \dfrac{\pi}{6}$

Separate: $\cot y\,dy = \cos x\,dx$.
Integrate: $\ln|\sin y| = \sin x + C$.
IC: $\ln|\sin\frac{\pi}{6}| = \sin\frac{3\pi}{2} + C \Rightarrow \ln\frac{1}{2} = -1 + C \Rightarrow C = \ln\frac{1}{2}+1$.
$\boxed{\ln|\sin y| = \sin x + \ln\tfrac{1}{2} + 1}$

7. Continuous Compound Interest A = A₀eᵏᵗ

7

Initial deposit \$500. After 5 years: \$600. How much after 18 years? Round to nearest cent.

$k = \frac{\ln(6/5)}{5}$.
$A(18) = 500e^{18k} = 500\cdot\left(\frac{6}{5}\right)^{18/5} \approx \boxed{\$963.88}$

8. Radioactive Decay Half-life

8

90% of a sample decays in 9 days. Find the half-life. Round to 2 decimal places.

$e^{9k} = \frac{1}{10} \Rightarrow k = \frac{\ln(1/10)}{9}$.
Half-life: $t_{1/2} = \frac{\ln(1/2)}{k} = \frac{9\ln 2}{\ln 10} \approx \boxed{2.71 \text{ days}}$

9. Parametric Curves — Sketching Parametric

9(a)

Sketch $x = -t^2,\ y = t^4+1,\ -\infty < t < \infty$. Describe the shape.

Note $x = -t^2 \le 0$ always, and $y = t^4+1 = x^2+1$.
Curve is the parabola $y = x^2+1$ for $x \le 0$, with vertex at $(0,1)$. As $t \to \pm\infty$, $x \to -\infty$ and $y \to +\infty$. Opens left along the parabola.

9(b)

Sketch $x = 4\cos 2t,\ y = 3\sin 2t,\ \frac{\pi}{2} \le t \le \pi$. Describe the shape and range.

Ellipse equation: $\frac{x^2}{16}+\frac{y^2}{9}=1$.
At $t=\pi/2$: $(x,y)=(-4,0)$. At $t=\pi$: $(x,y)=(4,0)$.
As $t$ goes $\pi/2 \to \pi$, $2t$ goes $\pi \to 2\pi$ — traces the bottom half of the ellipse from $(-4,0)$ to $(4,0)$.

10. Parametric Derivatives dy/dx and d²y/dx²

10(a)

Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for $x = t^3+3t,\ y = 3t^4+6t^2$.

$\frac{dx}{dt} = 3t^2+3,\quad \frac{dy}{dt} = 12t^3+12t$
$\frac{dy}{dx} = \frac{12t(t^2+1)}{3(t^2+1)} = 4t$
$\frac{d^2y}{dx^2} = \frac{d(4t)/dt}{dx/dt} = \frac{4}{3(t^2+1)}$

10(b)

Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for $x = 5\cos t,\ y = 3\sin t$.

$\frac{dy}{dx} = \frac{3\cos t}{-5\sin t} = -\frac{3}{5}\cot t$
$\frac{d^2y}{dx^2} = \frac{\frac{3}{5}\csc^2 t}{-5\sin t} = -\frac{3}{25\sin^3 t}$

11. Polar Curves — Symmetry Polar

11(a) — $r = 3 + 2\cos\theta$

Sketch and test symmetry about x-axis, y-axis, and origin.

x-axis ($\theta \to -\theta$): $\cos(-\theta)=\cos\theta$ → equation unchanged. Symmetric ✓

y-axis ($\theta \to \pi-\theta$): $\cos(\pi-\theta)=-\cos\theta$ → $r = 3-2\cos\theta$. Not identical. Not proven.

Origin ($r \to -r$): $-r = 3+2\cos\theta$. Not identical. Not proven.

Shape: limaçon (no inner loop), widest at $\theta=0$ ($r=5$), narrowest at $\theta=\pi$ ($r=1$).

11(b) — $r = 1 - 2\sin\theta$

Sketch and test symmetry about x-axis, y-axis, and origin.

x-axis ($\theta \to -\theta$): $\sin(-\theta)=-\sin\theta$ → $r = 1+2\sin\theta$. Not identical. Not proven.

y-axis ($\theta \to \pi-\theta$): $\sin(\pi-\theta)=\sin\theta$ → equation unchanged. Symmetric ✓

Origin ($r \to -r$): Not identical. Not proven.

Shape: limaçon with inner loop (since $|{-2}| > 1$). Inner loop when $1 - 2\sin\theta < 0$, i.e. $\sin\theta > 1/2$.

Calc 2 — Full Practice Test