Section 2.8 Practice Gateway Exam 8
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.
\(\displaystyle \int \left( 3x^4 + \frac{1}{\sqrt{x}} + 2^{3x}\right) \, dx \)
Solution.
\begin{align*}
\int \left( 3x^4 + \frac{1}{\sqrt{x}} + 2^{3x}\right) \, dx \amp = \int (3x^4 + x^{-1/2} + 2^{3x}) \, dx\\
\amp = 3\cdot \frac{x^5}{5} + \frac{x^{1/2}}{1/2} + \frac{1}{3} \cdot \frac{1}{\ln(2)}2^{3x} + C\\
\amp = \frac{3}{5} x^5 + 2\sqrt{x} + \frac{2^{3x}}{3\ln(2)} + C
\end{align*}
2.
\(\displaystyle \int t^2e^{2t} \, dt \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = t^2 \amp dv \amp = e^{2t} \, dt\\
du \amp = 2t \, dt \amp v \amp = \frac{1}{2} e^{2t}
\end{align*}
Then
\begin{align*}
\int t^2e^{2t} \, dt \amp = t^2\cdot \frac{1}{2}e^{2t} - \int \frac{1}{2}e^{2t} \cdot 2t \, dt\\
\amp =\frac{1}{2}t^2 e^{2t} - \int te^{2t} \, dt
\end{align*}
We now use integration by parts again on the new integral.
\begin{align*}
u \amp = t \amp dv \amp = e^{2t} \, dt\\
du \amp = 1\, dt \amp v \amp = \frac{1}{2}e^{2t}
\end{align*}
So
\begin{align*}
\int t^2e^{2t} \, dt \amp = \frac{1}{2}t^2 e^{2t} - \left(t\cdot \frac{1}{2}e^{2t} - \int \frac{1}{2} e^{2t} \, dt\right)\\
\amp = \frac{1}{2}t^2 e^{2t} -\frac{1}{2}te^{2t} + \frac{1}{2} \cdot \frac{1}{2} e^{2t} + C
\end{align*}
3.
\(\displaystyle \int 5x^2(x^3+4)^{12} \, dx \)
Solution.
Use substitution with \(u=x^3 + 4\) and \(du = 3x^2 \, dx\text{,}\) so \(\dfrac{1}{3} \, du = x^2 \, dx\)
\begin{align*}
\int 5x^2(x^3+4)^{12} \, dx \amp = \int 5 \cdot \frac{1}{3} u^{12} \, du\\
\amp = \frac{5}{3} \cdot \frac{u^{13}}{13} + C\\
\amp = \frac{5}{39} (x^3+4)^{13} + C
\end{align*}
4.
\(\displaystyle \int \frac{x}{(x-2)(x-3)} \, dx \)
Solution.
First decompose the integrand into partial fractions:
\begin{equation*}
\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}
\end{equation*}
This yields
\begin{equation*}
x = 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*}
2 = -A + 0 \implies A=-2
\end{equation*}
\begin{equation*}
3 = 0 + B \implies B=3
\end{equation*}
Now perform the integration:
\begin{align*}
\int \frac{x}{(x-2)(x-3)} \, dx \amp = \int \left(\frac{-2}{x-2} + \frac{3}{x-3}\right)\, dx\\
\amp = -2\ln|x-2| + 3\ln|x-3| + C
\end{align*}
5.
\(\displaystyle \int \frac{\sin(\theta)}{\cos(\theta)} \, d\theta \)
Solution.
Use substitution with \(u=\cos(\theta)\) and \(du = -\sin(\theta) \, d\theta\text{,}\) so \(-du = \sin(\theta) \, d\theta\text{:}\)
\begin{align*}
\int \frac{\sin(\theta)}{\cos(\theta)} \, d\theta \amp = \int - \frac{1}{u} \, du\\
\amp = -\ln|u| + C\\
\amp = -\ln|\cos(\theta)| + C
\end{align*}
6.
\(\displaystyle \int x^8 \ln(x) \, dx \)
Solution.
Use integration by parts with
\begin{align*}
u \amp = \ln(x) \amp dv \amp = x^8 \, dx\\
du \amp = \dfrac{1}{x} \, dx \amp v \amp = \dfrac{x^9}{9}
\end{align*}
Then
\begin{align*}
\int x^8 \ln(x) \, dx \amp = \ln(x) \cdot \frac{x^9}{9} - \int \frac{x^9}{9} \cdot \frac{1}{x} \, dx\\
\amp = \frac{1}{9}x^9 \ln(x) - \int \frac{1}{9} x^8 \, dx\\
\amp = \frac{1}{9} x^9 \ln(x) - \frac{1}{9}\cdot \frac{1}{9} x^9 + C
\end{align*}
7.
\(\displaystyle \int \frac{3x+2}{6x^2+8x+5} \, dx \)
Solution.
Use substitution with \(u=6x^2+8x+5\) and \(du = (12x + 8) \, dx\text{,}\) so \(\dfrac{1}{4} \, du = (3x+2)\, dx\text{:}\)
\begin{align*}
\int \frac{3x+2}{6x^2+8x+5} \, dx \amp = \int \frac{1}{4}\cdot \frac{1}{u} \, du\\
\amp = \frac{1}{4}\ln|u| + C\\
\amp = \frac{1}{4}\ln|6x^2 + 8x + 5| + C
\end{align*}
8.
\(\displaystyle \int 7t\cos(t^2+5) \, dt \)
Solution.
Use substitution with \(u = t^2 + 5\) and \(du = 2t \, dt\text{,}\) so \(\dfrac{1}{2} \, du = t \, dt\text{:}\)
\begin{align*}
\int 7t\cos(t^2+5) \, dt \amp = \int 7 \cdot \frac{1}{2} \cos(u) \, du\\
\amp = \frac{7}{2} \sin(u) + C\\
\amp = \frac{7}{2} \sin(t^2+5) + C
\end{align*}
