Section 2.3 Practice Gateway Exam 3
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.
\(\displaystyle \int \left(\frac{1}{2}x^2 - 2\sqrt{x} +3\right) \, dx \)
Solution.
\begin{align*}
\int \left(\frac{1}{2}x^2 - 2\sqrt{x} +3\right) \, dx \amp = \int \left( \frac{1}{2} x^2 - 2x^{1}{2} + 3\right) \, dx\\
\amp = \frac{1}{2} \cdot \frac{x^3}{3} - 2\frac{x^{3/2}}{3/2} + 3x + C\\
\amp = \frac{x^3}{6} - \frac{4}{3} x^{3/2} + 3x + C
\end{align*}
2.
\(\displaystyle \int \frac{5t}{t^2+5} \, dt \)
Solution.
Use substitution with \(u=t^2+5\) and \(du = 2t \, dt\text{,}\) so \(\dfrac{1}{2} \, du = t \, dt\text{:}\)
\begin{align*}
\int \frac{5t}{t^2+5} \, dt \amp = \int 5 \cdot \frac{1}{2} \frac{1}{u} \, du\\
\amp = \frac{5}{2} \ln|u| + C\\
\amp = \frac{5}{2} \ln|t^2+5| + C
\end{align*}
3.
\(\displaystyle \int xe^{2x^2-3} \, dx \)
Solution.
Use substitution with \(u=2x^2-3\) and \(du=4x \, dx\text{,}\) so \(\dfrac{1}{4} \, du = x \, dx\text{:}\)
\begin{align*}
\int xe^{2x^2-3} \, dx \amp = \int \frac{1}{4} e^u \, du\\
\amp = \frac{1}{4} e^u + C\\
\amp = \frac{1}{4} e^{2x^2-3} + C
\end{align*}
4.
\(\displaystyle \int \frac{7}{(x-3)(x+7)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{7}{(x-3)(x+7)} = \frac{A}{x-3} + \frac{B}{x+7}
\end{equation*}
This yields
\begin{equation*}
7 = A(x+7) + B(x-3)\text{.}
\end{equation*}
Plug in \(x=3\) to solve for \(A\) and \(x=-7\) to solve for \(B\text{:}\)
\begin{equation*}
7 = 10A + 0 \implies A=\frac{7}{10}
\end{equation*}
\begin{equation*}
7 = 0 - 10B \implies B=-\frac{7}{10}
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{7}{(x-3)(x+7)} \, dx \amp = \int \left( \frac{7}{10}\cdot \frac{1}{x-3} - \frac{7}{10} \cdot\frac{1}{x+7} \right) \, dx\\
\amp = \frac{7}{10} \ln|x-3| - \frac{7}{10} \ln|x+7| + C
\end{align*}
5.
\(\displaystyle \int \frac{2x^3+3x}{x^2} \, dx \)
Solution.
\begin{align*}
\int \frac{2x^3+3x}{x^2} \, dx \amp = \int (2x + 3x^{-1}) \, dx\\
\amp = 2\cdot \frac{x^2}{2} + 3\ln|x| + C\\
\amp = x^2 + 3 \ln|x| + C
\end{align*}
6.
\(\displaystyle \int 2xe^{3x} \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = 2x \amp dv \amp = e^{3x} \, dx\\
du \amp = 2 \, dx \amp v \amp = \frac{1}{3}e^x
\end{align*}
Then
\begin{align*}
\int 2xe^{3x} \, dx \amp = 2x \cdot \frac{1}{3}e^{3x} - \int \frac{1}{3} e^{3x} \cdot 2 \, dx\\
\amp = \frac{2}{3} x e^{3x} - \frac{2}{9} e^{3x} + C
\end{align*}
7.
\(\displaystyle \int 2x^3 \ln(x) \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = \ln(x) \amp dv \amp = 2x^3 \, dx\\
du \amp = \frac{1}{x} \, dx \amp v \amp = \frac{1}{2}x^4
\end{align*}
Then
\begin{align*}
\int 2x^3 \ln(x) \, dx \amp = \ln(x) \cdot \frac{1}{2} x^4 - \int \frac{1}{2} x^4 \cdot \frac{1}{x} \, dx\\
\amp = \frac{1}{2} x^4 \ln(x) - \frac{1}{2} \int x^3 \, dx\\
\amp = \frac{1}{2} x^4 \ln(x) -\frac{1}{2} \cdot \frac{x^4}{4} + C
\end{align*}
8.
\(\displaystyle \int x^2(5x^3-8)^4 \, dx \)
Solution.
Use substitution with \(u=5x^3-8\) and \(du=15x^2 \, dx\text{,}\) so \(\dfrac{1}{15} \, du = x^2 \, dx\text{:}\)
\begin{align*}
\int x^2(5x^3-8)^4 \, dx \amp = \int \frac{1}{15} \cdot u^4 \, du\\
\amp = \frac{1}{15} \cdot \frac{u^5}{5} + C\\
\amp = \frac{(5x^3-8)^5}{75} + C
\end{align*}
