Section 2.7 Practice Gateway Exam 7
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.
\(\displaystyle \int \left(x^3+ \frac{1}{x} + e^{3x}\right) \, dx \)
Solution.
\begin{align*}
\int \left(x^3+ \frac{1}{x} + e^{3x}\right) \, dx \amp = \frac{x^4}{4} + \ln|x| + \frac{1}{3} e^{3x} + C\\
\end{align*}
2.
\(\displaystyle \int 2x\cos(3x) \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = 2x \amp dv \amp = \cos(3x)\, dx\\
du \amp = 2 \, dx \amp v \amp = \dfrac{1}{3}\sin(3x)
\end{align*}
Then
\begin{align*}
\int 2x\cos(3x) \, dx \amp = 2x\cdot \frac{1}{3} \sin(3x) - \int 2 \cdot \frac{1}{3} \sin(3x) \, dx\\
\amp = \frac{2}{3}x\sin(3x) - \frac{2}{3} \cdot \frac{1}{3}\cdot -\cos(3x) + C\\
\amp = \frac{2}{3}x\sin(3x) +\frac{2}{9} \cos(3x) + C
\end{align*}
3.
\(\displaystyle \int \frac{2x^5+7x}{x^3} \, dx \)
Solution.
\begin{align*}
\int \frac{2x^5+7x}{x^3} \, dx \amp = \int (2x^2 + 7x^{-2}) \, dx\\
\amp = 2\cdot \frac{x^3}{3} + 7 \cdot \frac{x^{-1}}{-1} + C\\
\amp = \frac{2}{3}x^3 -7x^{-1} + C
\end{align*}
4.
\(\displaystyle \int \frac{8}{(x-2)(x+3)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{8}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}
\end{equation*}
This yields
\begin{equation*}
8 = 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*}
8 = 5A + 0 \implies A=\frac{8}{5}
\end{equation*}
\begin{equation*}
8 = 0 - 5B \implies B=-\frac{8}{5}
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{8}{(x-2)(x+3)} \, dx \amp = \int \left(\frac{8}{5} \cdot \frac{1}{x-2} - \frac{8}{5} \cdot \frac{1}{x+3}\right)\\
\amp = \frac{8}{5}\ln|x-2| - \frac{8}{5} \ln|x+3| + C
\end{align*}
5.
\(\displaystyle \int \sin(t)\cos^4(t) \, dt \)
Solution.
Use substitution with \(u=\cos(t)\) and \(du = -\sin(t) \, dt\text{,}\) so \(-du = \sin(t) \, dt\text{:}\)
\begin{align*}
\int \sin(t)\cos^4(t) \, dt \amp = \int -u^4 \, du\\
\amp = -\frac{u^5}{5} + C\\
\amp = -\frac{1}{5} \cos^5(t) + C
\end{align*}
6.
\(\displaystyle \int x\ln(4x) \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = \ln(4x) \amp dv \amp = x \, dx\\
du \amp = \dfrac{1}{x} \, dx \amp v \amp = \dfrac{x^2}{2}
\end{align*}
Then
\begin{align*}
\int x\ln(4x) \, dx \amp = \ln(4x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx\\
\amp = \frac{1}{2} x^2\ln(4x) - \int \frac{1}{2}x \, dx\\
\amp = \frac{1}{2} x^2 \ln(4x) - \frac{1}{2} \cdot \frac{x^2}{2} + C
\end{align*}
7.
\(\displaystyle \int \frac{x}{(121+x^2)^2} \, dx \)
Solution.
Use substitution with \(u=121+x^2\) and \(du=2x \, dx\text{,}\) so \(\dfrac{1}{2} \, du = x \, dx\text{:}\)
\begin{align*}
\int \frac{x}{(121+x^2)^2} \, dx \amp = \int \frac{1}{2} u^{-2} \, du\\
\amp = \frac{1}{2} \frac{u^{-1}}{-1} + C\\
\amp = -\frac{1}{2} \cdot (121+x^2)^{-1} + C
\end{align*}
8.
\(\displaystyle \int 3w\sin(5w^2+1) \, dw \)
Solution.
Use substitution with \(u=5w^2 + 1\) and \(du=10w \, dw\text{,}\) so \(\dfrac{1}{10} \, du = w \, dw\text{:}\)
\begin{align*}
\int 3w\sin(w^2+1) \, dw \amp = \int 3 \cdot \frac{1}{10} \sin(u) \, du\\
\amp = -\frac{3}{10} \cos(u) + C\\
\amp = -\frac{3}{10} \cos(5w^2 + 1) + C
\end{align*}
