Section 2.6 Practice Gateway Exam 6
The use of calculators or any computer algebra system is not permitted on this test. No partial credit will be awarded. A score of at least 7 correct is required to pass this gateway.
Compute each of the following integrals.
Exercises Exercises
1.
\(\displaystyle \int (y^{-3} + 8e) \, dy \)
Solution.
\begin{align*}
\int (y^{-3} + 8e) \, dy \amp = \frac{y^{-2}}{-2} + 8ey + C\\
\end{align*}
2.
\(\displaystyle \int xe^{2x} \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = x \amp dv \amp = e^{2x} \, dx\\
du \amp = 1 \, dx \amp v \amp = \dfrac{1}{2}e^{2x}
\end{align*}
Then
\begin{align*}
\int xe^{2x} \, dx \amp = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \, dx\\
\amp = \frac{1}{2}xe^{2x} - \frac{1}{2} \cdot \frac{1}{2}e^{2x} + C
\end{align*}
3.
\(\displaystyle \int \frac{x+3}{\sqrt{x^2+6x-11}} \, dx \)
Solution.
Use substitution with \(u= x^2+6x-11\) and \(du = (2x+6) \, dx\text{,}\) so \(\dfrac{1}{2} \, du = (x+3) \, dx\text{:}\)
\begin{align*}
\int \frac{x+3}{\sqrt{x^2+6x-11}} \, dx \amp = \int \frac{1}{2} \frac{1}{\sqrt{u}} \, du\\
\amp = \frac{1}{2} \int u^{-1/2} \, du\\
\amp = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} + C\\
\amp = \sqrt{x^2+6x-11} + C
\end{align*}
4.
\(\displaystyle \int z\cos(\pi z^2) \, dz \)
Solution.
Use substitution with \(u= \pi z^2\) and \(du = 2\pi z \, dz\text{,}\) so \(\dfrac{1}{2\pi} \, du = z \, dz\text{:}\)
\begin{align*}
\int z\cos(\pi z^2) \, dz \amp = \int \frac{1}{2\pi} \cos(u) \, du\\
\amp = \frac{1}{2\pi} \sin(u) + C\\
\amp = \frac{1}{2\pi} \sin(\pi z^2) + C
\end{align*}
5.
\(\displaystyle \int \sin(5\theta)\cos(5\theta) \, d\theta \)
Solution.
Use substitution with \(u = \sin(5\theta)\) and \(du = 5\cos(5\theta) \, d\theta\text{,}\) so \(\dfrac{1}{5} \, du = \cos(5\theta) \, d\theta\text{:}\)
\begin{align*}
\int \sin(5\theta)\cos(5\theta) \, d\theta \amp = \int \frac{1}{5} u \, du\\
\amp = \frac{1}{5} \frac{u^2}{2} + C\\
\amp = \frac{\sin^2(5\theta)}{10} + C
\end{align*}
6.
\(\displaystyle \int \frac{x}{1+9x^2} \, dx \)
Solution.
Use substitution with \(u = 1+9x^2\) and \(du = 18x \, dx\text{,}\) so \(\dfrac{1}{18} \, du = x \, dx\text{:}\)
\begin{align*}
\int \frac{x}{1+9x^2} \, dx \amp = \int \frac{1}{18} \frac{1}{u} \, du\\
\amp = \frac{1}{18} \ln|u| + C\\
\amp = \frac{1}{18} \ln|1+9x^2| + C
\end{align*}
7.
\(\displaystyle \int t^2 \ln(t) \, dt \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = \ln(t) \amp dv \amp = t^2 \, dt\\
du \amp = \frac{1}{t} \, dt \amp v \amp = \frac{t^3}{3}
\end{align*}
Then
\begin{align*}
\int t^2 \ln(t) \, dt \amp = \ln(t) \cdot \frac{t^3}{3} - \int \frac{t^3}{3} \cdot \frac{1}{t} \, dt\\
\amp = \frac{1}{3} t^3 \ln(t) - \int \frac{1}{3} t^2 \, dt\\
\amp = \frac{1}{3} t^3 \ln(t) - \frac{1}{3} \cdot \frac{t^3}{3} + C
\end{align*}
8.
\(\displaystyle \int \frac{1}{(x+2)(x-3)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}
\end{equation*}
This yields
\begin{equation*}
1 = A(x-3) + B(x+2)\text{.}
\end{equation*}
Plug in \(x=-2\) to solve for \(A\) and \(x=3\) to solve for \(B\text{:}\)
\begin{equation*}
1 = -5A + 0 \implies A=-\frac{1}{5}
\end{equation*}
\begin{equation*}
1 = 0 + 5B \implies B=\frac{1}{5}
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{1}{(x+2)(x-3)} \, dx \amp = \int \left(-\frac{1}{5}\cdot \frac{1}{x+2} + \frac{1}{5} \cdot \frac{1}{x-3}\right) \, dx\\
\amp = -\frac{1}{5} \ln|x+2| +\frac{1}{5} \ln|x-3| + C
\end{align*}
