Section 2.2 Practice Gateway Exam 2
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.
\(\displaystyle \int \left(3x^5 + \frac{1}{x^3} + 5\right) \, dx\)
Solution.
\begin{align*}
\int \left(3x^5 + \frac{1}{x^3} + 5\right) \, dx \amp = \int ( 3x^5 + x^{-3} + 5) \, dx\\
\amp = 3 \frac{x^6}{6} + \frac{x^{-2}}{2} + 5x + C
\end{align*}
2.
\(\displaystyle \int \frac{2x}{3x^2+4} \, dx \)
Solution.
Use substitution with \(u=3x^2+4\) and \(du=6x \, dx\text{,}\) so \(\dfrac{1}{6} du = x \, dx\text{:}\)
\begin{align*}
\int \frac{2x}{3x^2+4} \, dx \amp = \int 2 \frac{1}{6} \frac{1}{u} \, du\\
\amp = \frac{1}{3} \ln|u| + C\\
\amp = \frac{1}{3}\ln|3x^2+4| + C
\end{align*}
3.
\(\displaystyle \int x^2e^{2x^3} \, dx \)
Solution.
Use subsitution with \(u=2x^3\) and \(du = 6x^2 \, dx\text{,}\) so \(\dfrac{1}{6} \, du = x^2 \, dx\text{:}\)
\begin{align*}
\int x^2e^{2x^3} \, dx \amp = \int \frac{1}{6} e^u \, du\\
\amp = \frac{1}{6} e^u + C\\
\amp = \frac{1}{6}e^{2x^3}+C
\end{align*}
4.
\(\displaystyle \int \frac{10}{(x-2)(x+5)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{10}{(x-2)(x+5)} = \frac{A}{x-2} + \frac{B}{x+5}
\end{equation*}
This yields
\begin{equation*}
10 = A(x+5) + B(x-2)\text{.}
\end{equation*}
Plug in \(x=2\) to solve for \(A\) and \(x=-5\) to solve for \(B\text{:}\)
\begin{equation*}
10 = 7A + 0 \implies A=\frac{10}{7}
\end{equation*}
\begin{equation*}
10= 0 -7B \implies B=-\frac{10}{7}
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{10}{(x-2)(x+5)} \, dx \amp = \int \left( \frac{10}{7}\frac{1}{x-2} - \frac{10}{7} \frac{1}{x+5} \right) \, dx\\
\amp = \frac{10}{7}\ln|x-2| - \frac{10}{7} \ln|x+5| + C
\end{align*}
5.
\(\displaystyle \int \frac{x^3+x+2}{x^2} \, dx \)
Solution.
\begin{align*}
\int \frac{x^3+x+2}{x^2} \, dx \amp = \int (x + x^{-1} + 2x^{-2}) \, dx\\
\amp = \frac{x^2}{2} + \ln|x| + 2\frac{x^{-1}}{-1} + C\\
\amp = \frac{x^2}{2} + \ln|x| -\frac{2}{x} + C
\end{align*}
6.
\(\displaystyle \int t\sin(t) \, dt \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = t \amp dv \amp = \sin(t) \, dt\\
du \amp = dt \amp v \amp = -\cos(t)
\end{align*}
Then
\begin{align*}
\int t\sin(t) \, dt \amp = t\cdot (-\cos(t)) - \int -\cos(t) \, dt\\
\amp = -t\cos(t)+ \int \cos(t) \, dt\\
\amp = -t\cos(t) + \sin(t) + C
\end{align*}
7.
\(\displaystyle \int 5x^2\cos(2x^3) \, dx \)
Solution.
Use substitution with \(u=2x^3\) and \(du = 6x^2 \, dx\text{,}\) so \(\dfrac{1}{6} \, du = x^2 \, dx\text{:}\)
\begin{align*}
\int 5x^2\cos(2x^3) \, dx \amp = \int 5 \frac{1}{6} \cos(u) \, du\\
\amp = \frac{5}{6} \sin(u) + C\\
\amp = \frac{5}{6} \sin(2x^3) + C
\end{align*}
8.
\(\displaystyle \int \frac{1}{\sqrt{z+8}} \, dz \)
Solution.
Use substitution with \(u=z+8\) and \(du=dz\text{:}\)
\begin{align*}
\int \frac{1}{\sqrt{z+8}} \, dz \amp = \int \frac{1}{u} \, du\\
\amp = \int u^{-1/2} \, du\\
\amp = \frac{u^{1/2}}{1/2} + C\\
\amp = 2(z+8)^{1/2} + C
\end{align*}
