Practice Gateway Exam 7

Section 2.7 Practice Gateway Exam 7

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.

$\int (x^{3} + \frac{1}{x} + e^{3 x}) d x$

Solution.

\begin{aligned} \int (x^{3} + \frac{1}{x} + e^{3 x}) d x & = \frac{x^{4}}{4} + \ln | x | + \frac{1}{3} e^{3 x} + C \end{aligned}

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2.

$\int 2 x \cos (3 x) d x$

Solution.

Use integration by parts with

\begin{aligned} u & = 2 x & d v & = \cos (3 x) d x \\ d u & = 2 d x & v & = \frac{1}{3} \sin (3 x) \end{aligned}

Then

\begin{aligned} \int 2 x \cos (3 x) d x & = 2 x \cdot \frac{1}{3} \sin (3 x) - \int 2 \cdot \frac{1}{3} \sin (3 x) d x \\ = \frac{2}{3} x \sin (3 x) - \frac{2}{3} \cdot \frac{1}{3} \cdot - \cos (3 x) + C \\ = \frac{2}{3} x \sin (3 x) + \frac{2}{9} \cos (3 x) + C \end{aligned}

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3.

$\int \frac{2 x^{5} + 7 x}{x^{3}} d x$

Solution.

\begin{aligned} \int \frac{2 x^{5} + 7 x}{x^{3}} d x & = \int (2 x^{2} + 7 x^{- 2}) d x \\ = 2 \cdot \frac{x^{3}}{3} + 7 \cdot \frac{x^{- 1}}{- 1} + C \\ = \frac{2}{3} x^{3} - 7 x^{- 1} + C \end{aligned}

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4.

$\int \frac{8}{(x - 2) (x + 3)} d x$

Solution.

First decompose the integrand into partial fractions:

\frac{8}{(x - 2) (x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}

This yields

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

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

8 = 5 A + 0 ⟹ A = \frac{8}{5}

8 = 0 - 5 B ⟹ B = - \frac{8}{5}

Now perform the integration:

\begin{aligned} \int \frac{8}{(x - 2) (x + 3)} d x & = \int (\frac{8}{5} \cdot \frac{1}{x - 2} - \frac{8}{5} \cdot \frac{1}{x + 3}) \\ = \frac{8}{5} \ln | x - 2 | - \frac{8}{5} \ln | x + 3 | + C \end{aligned}

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5.

$\int \sin (t) \cos^{4} (t) d t$

Solution.

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

\begin{aligned} \int \sin (t) \cos^{4} (t) d t & = \int - u^{4} d u \\ = - \frac{u^{5}}{5} + C \\ = - \frac{1}{5} \cos^{5} (t) + C \end{aligned}

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6.

$\int x \ln (4 x) d x$

Solution.

Use integration by parts with

\begin{aligned} u & = \ln (4 x) & d v & = x d x \\ d u & = \frac{1}{x} d x & v & = \frac{x^{2}}{2} \end{aligned}

Then

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

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7.

$\int \frac{x}{(121 + x^{2})^{2}} d x$

Solution.

Use substitution with $u = 121 + x^{2}$ and $d u = 2 x d x,$ so $\frac{1}{2} d u = x d x :$

\begin{aligned} \int \frac{x}{(121 + x^{2})^{2}} d x & = \int \frac{1}{2} u^{- 2} d u \\ = \frac{1}{2} \frac{u^{- 1}}{- 1} + C \\ = - \frac{1}{2} \cdot (121 + x^{2})^{- 1} + C \end{aligned}

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8.

$\int 3 w \sin (5 w^{2} + 1) d w$

Solution.

Use substitution with $u = 5 w^{2} + 1$ and $d u = 10 w d w,$ so $\frac{1}{10} d u = w d w :$

\begin{aligned} \int 3 w \sin (w^{2} + 1) d w & = \int 3 \cdot \frac{1}{10} \sin (u) d u \\ = - \frac{3}{10} \cos (u) + C \\ = - \frac{3}{10} \cos (5 w^{2} + 1) + C \end{aligned}