Practice Gateway Exam 8

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.

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Compute each of the following integrals.

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Exercises Exercises

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1.

$\int (3 x^{4} + \frac{1}{\sqrt{x}} + 2^{3 x}) d x$

Solution.

\begin{aligned} \int (3 x^{4} + \frac{1}{\sqrt{x}} + 2^{3 x}) d x & = \int (3 x^{4} + x^{- 1 / 2} + 2^{3 x}) d x \\ = 3 \cdot \frac{x^{5}}{5} + \frac{x^{1 / 2}}{1 / 2} + \frac{1}{3} \cdot \frac{1}{\ln (2)} 2^{3 x} + C \\ = \frac{3}{5} x^{5} + 2 \sqrt{x} + \frac{2^{3 x}}{3 \ln (2)} + C \end{aligned}

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2.

$\int t^{2} e^{2 t} d t$

Solution.

Use integration by parts with

\begin{aligned} u & = t^{2} & d v & = e^{2 t} d t \\ d u & = 2 t d t & v & = \frac{1}{2} e^{2 t} \end{aligned}

Then

\begin{aligned} \int t^{2} e^{2 t} d t & = t^{2} \cdot \frac{1}{2} e^{2 t} - \int \frac{1}{2} e^{2 t} \cdot 2 t d t \\ = \frac{1}{2} t^{2} e^{2 t} - \int t e^{2 t} d t \end{aligned}

We now use integration by parts again on the new integral.

\begin{aligned} u & = t & d v & = e^{2 t} d t \\ d u & = 1 d t & v & = \frac{1}{2} e^{2 t} \end{aligned}

\begin{aligned} \int t^{2} e^{2 t} d t & = \frac{1}{2} t^{2} e^{2 t} - (t \cdot \frac{1}{2} e^{2 t} - \int \frac{1}{2} e^{2 t} d t) \\ = \frac{1}{2} t^{2} e^{2 t} - \frac{1}{2} t e^{2 t} + \frac{1}{2} \cdot \frac{1}{2} e^{2 t} + C \end{aligned}

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3.

$\int 5 x^{2} (x^{3} + 4)^{12} d x$

Solution.

Use substitution with $u = x^{3} + 4$ and $d u = 3 x^{2} d x,$ so $\frac{1}{3} d u = x^{2} d x$

\begin{aligned} \int 5 x^{2} (x^{3} + 4)^{12} d x & = \int 5 \cdot \frac{1}{3} u^{12} d u \\ = \frac{5}{3} \cdot \frac{u^{13}}{13} + C \\ = \frac{5}{39} (x^{3} + 4)^{13} + C \end{aligned}

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4.

$\int \frac{x}{(x - 2) (x - 3)} d x$

Solution.

First decompose the integrand into partial fractions:

\frac{x}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}

This yields

x = A (x - 3) + B (x - 2) .

Plug in $x = 2$ to solve for $A$ and $x = 3$ to solve for $B :$

2 = - A + 0 ⟹ A = - 2

3 = 0 + B ⟹ B = 3

Now perform the integration:

\begin{aligned} \int \frac{x}{(x - 2) (x - 3)} d x & = \int (\frac{- 2}{x - 2} + \frac{3}{x - 3}) d x \\ = - 2 \ln | x - 2 | + 3 \ln | x - 3 | + C \end{aligned}

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5.

$\int \frac{\sin (θ)}{\cos (θ)} d θ$

Solution.

Use substitution with $u = \cos (θ)$ and $d u = - \sin (θ) d θ,$ so $- d u = \sin (θ) d θ :$

\begin{aligned} \int \frac{\sin (θ)}{\cos (θ)} d θ & = \int - \frac{1}{u} d u \\ = - \ln | u | + C \\ = - \ln | \cos (θ) | + C \end{aligned}

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6.

$\int x^{8} \ln (x) d x$

Solution.

Use integration by parts with

\begin{aligned} u & = \ln (x) & d v & = x^{8} d x \\ d u & = \frac{1}{x} d x & v & = \frac{x^{9}}{9} \end{aligned}

Then

\begin{aligned} \int x^{8} \ln (x) d x & = \ln (x) \cdot \frac{x^{9}}{9} - \int \frac{x^{9}}{9} \cdot \frac{1}{x} d x \\ = \frac{1}{9} x^{9} \ln (x) - \int \frac{1}{9} x^{8} d x \\ = \frac{1}{9} x^{9} \ln (x) - \frac{1}{9} \cdot \frac{1}{9} x^{9} + C \end{aligned}

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7.

$\int \frac{3 x + 2}{6 x^{2} + 8 x + 5} d x$

Solution.

Use substitution with $u = 6 x^{2} + 8 x + 5$ and $d u = (12 x + 8) d x,$ so $\frac{1}{4} d u = (3 x + 2) d x :$

\begin{aligned} \int \frac{3 x + 2}{6 x^{2} + 8 x + 5} d x & = \int \frac{1}{4} \cdot \frac{1}{u} d u \\ = \frac{1}{4} \ln | u | + C \\ = \frac{1}{4} \ln | 6 x^{2} + 8 x + 5 | + C \end{aligned}

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8.

$\int 7 t \cos (t^{2} + 5) d t$

Solution.

Use substitution with $u = t^{2} + 5$ and $d u = 2 t d t,$ so $\frac{1}{2} d u = t d t :$

\begin{aligned} \int 7 t \cos (t^{2} + 5) d t & = \int 7 \cdot \frac{1}{2} \cos (u) d u \\ = \frac{7}{2} \sin (u) + C \\ = \frac{7}{2} \sin (t^{2} + 5) + C \end{aligned}