Section 2.1 Practice Gateway Exam 1
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.
\(\displaystyle \int (2x^3+x+12) \, dx\)
Solution.
\begin{align*}
\int (2x^3+x+12) \, dx \amp = 2\cdot \frac{x^4}{4} + \frac{x^2}{2} + 12x + C\\
\amp = \frac{1}{2}x^4 + \frac{1}{2} x^2 + 12x + C
\end{align*}
2.
\(\displaystyle \int \frac{4x}{2x^2+3} \, dx\)
Solution.
Use substitution with \(u=2x^2+3\) and \(du = 4x \, dx\text{.}\)
\begin{align*}
\int \frac{4x}{2x^2+3} \, dx \amp = \int \frac{1}{u} \, du\\
\amp = \ln|u| + C\\
\amp = \ln|2x^2+3|+C
\end{align*}
3.
\(\displaystyle \int 3e^{5z} \, dz\)
Solution.
Use substitution with \(u=5z\) and \(du = 5\, dz\text{,}\) so \(\dfrac{1}{5}\, du = dz\text{.}\)
\begin{align*}
\int 3e^{5z} \, dz \amp = \int 3e^u\cdot \frac{1}{5} \, du\\
\amp = \frac{3}{5} \int e^u \, du\\
\amp = \frac{3}{5}e^u + C\\
\amp = \frac{3}{5}e^{5z} + C
\end{align*}
4.
\(\displaystyle \int \frac{4}{x(x-2)}\, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{4}{x(x-2)} = \frac{A}{x} + \frac{B}{x-2}
\end{equation*}
This yields
\begin{equation*}
4 = A(x-2) + Bx
\end{equation*}
Plug in \(x=2\) to solve for \(B\) and \(x=0\) to solve for \(A\text{:}\)
\begin{equation*}
4 = 0 + 2B \implies B=2
\end{equation*}
\begin{equation*}
4=A(-2) + 0 \implies A=-2
\end{equation*}
Now perform the integration
\begin{align*}
\int \frac{4}{x(x-2)}\, dx \amp = \int\left(\frac{-2}{x} + \frac{2}{x-2}\right) \, dx\\
\amp = -2\ln|x| + 2\ln|x-2| + C
\end{align*}
5.
\(\displaystyle \int \left(3x^{-2} + \frac{1}{x}\right) \, dx\)
Solution.
\begin{align*}
\int \left(3x^{-2} + \frac{1}{x}\right) \, dx \amp = 3\frac{x^{-1}}{-1} + \ln|x| +C\\
\amp = -3 x^{-1} + \ln|x| + C
\end{align*}
6.
\(\displaystyle \int 8t \cos(t) \, dt\)
Solution.
Use integration by parts with
\begin{align*}
u \amp = 8t \amp dv \amp = \cos(t) \, dt\\
du \amp = 8 \, dt \amp v \amp = \sin(t)
\end{align*}
Then
\begin{align*}
\int 8t \cos(t) \amp = 8t\sin(t) - \int 8\sin(t) \, dt \\
\amp = 8t\sin(t) + 8\cos(t)+C
\end{align*}
7.
\(\displaystyle \int 2x^3(x^4+5)^6 \, dx\)
Solution.
Use substitution with \(u=x^4+5\) and \(du = 4x^3 \, dx\text{,}\) so \(\dfrac{1}{4} \, du = x^3 \, dx\text{.}\)
\begin{align*}
\int 2x^3(x^4+5)^6 \, dx \amp = 2\int u^6 \cdot \frac{1}{4} \, du\\
\amp = \frac{1}{2} \frac{u^7}{7} + C\\
\amp = \frac{(x^4+5)^7}{14} + C
\end{align*}
8.
\(\displaystyle \int \sqrt{4x+7} \, dx \)
Solution.
Use substitution with \(u=4x+7\) and \(du = 4\, dx\text{,}\) so \(\dfrac{1}{4} \, du = dx\text{.}\)
\begin{align*}
\int \sqrt{4x+7} \, dx \amp = \int u^{1/2} \cdot \frac{1}{4} \, du\\
\amp \frac{1}{4} \cdot \frac{u^{3/2}}{3/2} + C\\
\amp = \frac{1}{4}\cdot \frac{2}{3} (4x+7)^{3/2} +C
\end{align*}
