Section 2.4 Practice Gateway Exam 4
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.
\(\displaystyle \int (r+1)^3 \, dr \)
Solution.
Use substitution with \(u=r+1\) and \(du=dr\text{:}\)
\begin{align*}
\int (r+1)^3 \, dr \amp = \int u^3 \, du\\
\amp = \frac{u^4}{4}+C\\
\amp = \frac{(r+1)^4}{4}+C
\end{align*}
2.
\(\displaystyle \int x^2e^{2x} \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = x^2 \amp dv \amp = e^{2x} \, dx\\
du \amp = 2x \, dx \amp v \amp = \frac{1}{2} e^{2x}
\end{align*}
Then
\begin{align*}
\int x^2e^{2x} \, dx \amp = x^2\cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \cdot 2x \, dx\\
\amp = \frac{1}{2}x^2 e^{2x} - \int xe^{2x} \, dx
\end{align*}
We now use integration by parts again on the new integral.
\begin{align*}
u \amp = x \amp dv \amp = e^{2x} \, dx\\
du \amp = dx \amp v \amp = \frac{1}{2} e^{2x}
\end{align*}
So
\begin{align*}
\int x^2e^{2x} \, dx \amp = \frac{1}{2}x^2 e^{2x} - \left( x\cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx\right)\\
\amp = \frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} +\frac{1}{4}e^{2x} + C
\end{align*}
3.
\(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx \)
Solution.
Use substitution with \(u=\sqrt{x}=x^{1/2}\) and \(du = \dfrac{1}{2\sqrt{x}} \, dx\text{,}\) so \(2 \, du = \dfrac{1}{\sqrt{x}} \, dx\text{:}\)
\begin{align*}
\int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx \amp = \int 2 \cos(u) \, du\\
\amp = 2\sin(u) + C\\
\amp = 2\sin(\sqrt{x}) + C
\end{align*}
4.
\(\displaystyle \int \frac{7x}{(x-2)(x+3)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{7x}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}
\end{equation*}
This yields
\begin{equation*}
7x = 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*}
14 = 5A + 0 \implies A=\frac{14}{5}
\end{equation*}
\begin{equation*}
-21 = 0 - 5B \implies B=\frac{21}{5}
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{7x}{(x-2)(x+3)} \, dx \amp = \int \left( \frac{14}{5} \cdot \frac{1}{x-2} + \frac{21}{5} \cdot \frac{1}{x+3}\right) \, dx\\
\amp = \frac{14}{5}\ln|x-2| + \frac{21}{5} \ln|x+3| + C
\end{align*}
5.
\(\displaystyle \int (e^x+3^x) \, dx \)
Solution.
\begin{align*}
\int (e^x+3^x) \, dx \amp = e^x + \frac{1}{\ln(3)}\cdot 3^x +C
\end{align*}
6.
\(\displaystyle \int \frac{1}{x\ln(x)} \, dx \)
Solution.
Use substitution with \(u=\ln(x)\) and \(du=\dfrac{1}{x} \, dx\text{:}\)
\begin{align*}
\int \frac{1}{x\ln(x)} \, dx \amp = \int \frac{1}{u} \, du\\
\amp = \ln|u| + C\\
\amp = \ln|\ln(x)|+C
\end{align*}
7.
\(\displaystyle \int \sin^5(2\theta)\cos(2\theta) \, d\theta \)
Solution.
Use substitution with \(u=\sin(2\theta)\) and \(du = 2\cos(2\theta) \, d\theta\text{,}\) so \(\dfrac{1}{2} \, du = \cos(2\theta) \, d\theta\text{:}\)
\begin{align*}
\int \sin^5(2\theta)\cos(2\theta) \, d\theta \amp = \int \frac{1}{2} u^5 \, du\\
\amp = \frac{1}{2} \frac{u^6}{6} + C\\
\amp = \frac{\sin^6(2\theta)}{12} + C
\end{align*}
8.
\(\displaystyle \int \frac{2x^2}{(9+x^3)^2} \, dx \)
Solution.
Use substitution with \(u=9+x^3\) and \(du = 3x^2 \, dx\text{,}\) so \(\dfrac{1}{3} \, du = x^2 \, dx\text{:}\)
\begin{align*}
\int \frac{2x^2}{(9+x^3)^2} \, dx \amp = \int \frac{2}{3} u^{-2} \, du\\
\amp = \frac{2}{3}\frac{u^{-1}}{-1} + C\\
\amp = -\frac{2}{3}\cdot \frac{1}{9+x^3} + C
\end{align*}
