E078 Section 2 — Practice Test (15 Q)

Section 2 (Integration) practice test: 15 problems with final answers and condensed worked solutions. Companion to the E078 Section 2 flashcards. Not official WGU material.

Q1 (2.1). Evaluate $\sum_{i=1}^{20}(2i+3)$.

Answer: $480$ | Split: $2\sum i + \sum 3 = 2\cdot\frac{20\cdot 21}{2} + 20\cdot 3 = 420 + 60 = 480$.

Q2 (2.1). Use the sum formulas to evaluate $\sum_{i=1}^{10}(i^2 - 2i)$.

Answer: $275$ | $\sum_{i=1}^{10} i^2 = \frac{10\cdot 11\cdot 21}{6} = 385$ and $2\sum_{i=1}^{10} i = 2\cdot 55 = 110$ | $385 - 110 = 275$.

Q3 (2.1). For $f(x) = x^2$ on $[0,2]$ with $n = 4$, compute the left-endpoint approximation $L_4$ and the right-endpoint approximation $R_4$. Which one is an upper sum?

Answer: $L_4 = 1.75$, $R_4 = 3.75$ (upper) | $\Delta x = 0.5$; points $0, 0.5, 1, 1.5, 2$ give values $0, 0.25, 1, 2.25, 4$ | $L_4 = 0.5(0+0.25+1+2.25) = 1.75$; $R_4 = 0.5(0.25+1+2.25+4) = 3.75$ | $f$ is increasing, so the right endpoint is the max on each strip: $R_4$ is the upper sum (true value $\frac{8}{3} \approx 2.67$ lies between).

Q4 (2.2). Use geometry (no antiderivatives) to find (a) the net signed area $\int_0^6 (x-2)\,dx$ and (b) the total area between $y = x - 2$ and the $x$-axis on $[0,6]$.

Answer: (a) $6$ (b) $10$ | Zero at $x = 2$. On $[0,2]$ the triangle is below the axis: base $2$, height $2$, area $2$ | On $[2,6]$ it is above: base $4$, height $4$, area $8$ | (a) Net $= 8 - 2 = 6$; (b) Total $= 8 + 2 = 10$.

Q5 (2.2/2.3). Find the average value of $f(x) = 3x^2 - 2x$ on $[1,3]$, then find the value $c$ in $[1,3]$ guaranteed by the Mean Value Theorem for Integrals.

Answer: $f_{\text{ave}} = 9$, $c = \frac{1+2\sqrt{7}}{3} \approx 2.10$ | $\int_1^3 (3x^2 - 2x)\,dx = \left[x^3 - x^2\right]_1^3 = (27-9) - (1-1) = 18$, so $f_{\text{ave}} = \frac{18}{3-1} = 9$ | Solve $3c^2 - 2c = 9$: $c = \frac{2 \pm \sqrt{4+108}}{6} = \frac{1 \pm 2\sqrt{7}}{3}$ | Only $c = \frac{1+2\sqrt{7}}{3} \approx 2.10$ lies in $[1,3]$.

Q6 (2.3). Find $\dfrac{d}{dx}\int_1^x \sqrt{t^3+1}\,dt$.

Answer: $\sqrt{x^3+1}$ | FTC Part 1: the upper limit is just $x$, so plug it in: $\frac{d}{dx}\int_1^x \sqrt{t^3+1}\,dt = \sqrt{x^3+1}$.

Q7 (2.3). Evaluate $\int_1^4 \left(3\sqrt{x} - \dfrac{2}{x^2}\right) dx$.

Answer: $\frac{25}{2}$ | Rewrite as $3x^{1/2} - 2x^{-2}$; antiderivative $2x^{3/2} + 2x^{-1} = 2x^{3/2} + \frac{2}{x}$ | $\left[2x^{3/2} + \frac{2}{x}\right]_1^4 = \left(2\cdot 8 + \frac{1}{2}\right) - (2+2) = 16.5 - 4 = 12.5 = \frac{25}{2}$.

Q8 (2.5). Find $\int x^2 (x^3+5)^4\,dx$.

Answer: $\frac{(x^3+5)^5}{15} + C$ | Let $u = x^3 + 5$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$ | $\frac{1}{3}\int u^4\,du = \frac{1}{3}\cdot\frac{u^5}{5} + C = \frac{(x^3+5)^5}{15} + C$.

Q9 (2.4). Evaluate $\int_{-2}^{2} \left(x^5 - 3x^3 + 4x^2 + 1\right) dx$ using even/odd symmetry.

Answer: $\frac{76}{3}$ | $x^5$ and $-3x^3$ are odd, so they integrate to $0$ on $[-2,2]$ | $4x^2 + 1$ is even: $2\int_0^2 (4x^2+1)\,dx = 2\left[\frac{4}{3}x^3 + x\right]_0^2 = 2\left(\frac{32}{3} + 2\right) = 2\cdot\frac{38}{3} = \frac{76}{3}$.

Q10 (2.4). A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \le t \le 4$. Find (a) the displacement and (b) the total distance traveled.

Answer: (a) $\frac{4}{3}$ m (b) $4$ m | Antiderivative $F(t) = \frac{t^3}{3} - 2t^2 + 3t$ | (a) Displacement $= F(4) - F(0) = \frac{64}{3} - 32 + 12 = \frac{4}{3}$ m | (b) $v = (t-1)(t-3) = 0$ at $t = 1, 3$; $F(0) = 0$, $F(1) = \frac{4}{3}$, $F(3) = 0$, $F(4) = \frac{4}{3}$ | Distance $= |F(1)-F(0)| + |F(3)-F(1)| + |F(4)-F(3)| = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ m.

Q11 (2.5). Evaluate $\int_0^{\pi/3} \sin x \cos^2 x\,dx$.

Answer: $\frac{7}{24}$ | $u = \cos x$, $du = -\sin x\,dx$; limits $x = 0 \to u = 1$, $x = \pi/3 \to u = \frac{1}{2}$ | $\int_1^{1/2} -u^2\,du = \int_{1/2}^{1} u^2\,du = \left.\frac{u^3}{3}\right|_{1/2}^{1} = \frac{1}{3} - \frac{1}{24} = \frac{7}{24}$.

Q12 (2.6). (a) Find $\int 5^{2x}\,dx$. (b) Evaluate $\int_1^{e^2} \dfrac{(\ln x)^2}{x}\,dx$.

Answer: (a) $\frac{5^{2x}}{2\ln 5} + C$ (b) $\frac{8}{3}$ | (a) $u = 2x$, $du = 2\,dx$: $\frac{1}{2}\int 5^u\,du = \frac{1}{2}\cdot\frac{5^u}{\ln 5} + C = \frac{5^{2x}}{2\ln 5} + C$ | (b) $u = \ln x$, $du = \frac{1}{x}dx$; limits $x = 1 \to u = 0$, $x = e^2 \to u = 2$: $\int_0^2 u^2\,du = \frac{8}{3}$.

Q13 (2.7). Evaluate (a) $\int_0^{3/2} \dfrac{dx}{\sqrt{9 - x^2}}$ and (b) $\int_0^1 \dfrac{x}{1 + x^4}\,dx$.

Answer: (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{8}$ | (a) $a = 3$: $\left.\sin^{-1}\left(\frac{x}{3}\right)\right|_0^{3/2} = \sin^{-1}\left(\frac{1}{2}\right) - 0 = \frac{\pi}{6}$ | (b) $u = x^2$, $du = 2x\,dx$; limits $0 \to 0$, $1 \to 1$: $\frac{1}{2}\int_0^1 \frac{du}{1+u^2} = \frac{1}{2}\left.\tan^{-1} u\right|_0^1 = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$.

Q14 (2.3). Let $F(x) = \int_{2x}^{x^3} \dfrac{1}{1+t^2}\,dt$. Find $F'(x)$.

Answer: $F'(x) = \frac{3x^2}{1+x^6} - \frac{2}{1+4x^2}$ | $F'(x) = f(\text{top})\cdot(\text{top})' - f(\text{bottom})\cdot(\text{bottom})'$ with $f(t) = \frac{1}{1+t^2}$ | $F'(x) = \frac{1}{1+(x^3)^2}\cdot 3x^2 - \frac{1}{1+(2x)^2}\cdot 2 = \frac{3x^2}{1+x^6} - \frac{2}{1+4x^2}$.

Q15 (2.1/2.2). Use the definition of the definite integral (limit of right-endpoint Riemann sums) to evaluate $\int_0^2 x^2\,dx$. Check your answer with FTC Part 2.

Answer: $\frac{8}{3}$ | $\Delta x = \frac{2}{n}$, right endpoints $x_i = \frac{2i}{n}$, so $f(x_i) = \frac{4i^2}{n^2}$ | $R_n = \sum_{i=1}^{n} \frac{4i^2}{n^2}\cdot\frac{2}{n} = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{4}{3}\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)$ | $\lim_{n\to\infty} R_n = \frac{4}{3}\cdot 1\cdot 2 = \frac{8}{3}$ | Check: $\left.\frac{x^3}{3}\right|_0^2 = \frac{8}{3}$.