Practice Gateway Exam 2

Section 2.2 Practice Gateway Exam 2

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.

$\int (3 x^{5} + \frac{1}{x^{3}} + 5) d x$

Solution.

\begin{aligned} \int (3 x^{5} + \frac{1}{x^{3}} + 5) d x & = \int (3 x^{5} + x^{- 3} + 5) d x \\ = 3 \frac{x^{6}}{6} + \frac{x^{- 2}}{2} + 5 x + C \end{aligned}

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2.

$\int \frac{2 x}{3 x^{2} + 4} d x$

Solution.

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

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

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3.

$\int x^{2} e^{2 x^{3}} d x$

Solution.

Use subsitution with $u = 2 x^{3}$ and $d u = 6 x^{2} d x,$ so $\frac{1}{6} d u = x^{2} d x :$

\begin{aligned} \int x^{2} e^{2 x^{3}} d x & = \int \frac{1}{6} e^{u} d u \\ = \frac{1}{6} e^{u} + C \\ = \frac{1}{6} e^{2 x^{3}} + C \end{aligned}

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4.

$\int \frac{10}{(x - 2) (x + 5)} d x$

Solution.

First decompose the integrand into partial fractions:

\frac{10}{(x - 2) (x + 5)} = \frac{A}{x - 2} + \frac{B}{x + 5}

This yields

10 = A (x + 5) + B (x - 2) .

Plug in $x = 2$ to solve for $A$ and $x = - 5$ to solve for $B :$

10 = 7 A + 0 ⟹ A = \frac{10}{7}

10 = 0 - 7 B ⟹ B = - \frac{10}{7}

Now perform the integration:

\begin{aligned} \int \frac{10}{(x - 2) (x + 5)} d x & = \int (\frac{10}{7} \frac{1}{x - 2} - \frac{10}{7} \frac{1}{x + 5}) d x \\ = \frac{10}{7} \ln | x - 2 | - \frac{10}{7} \ln | x + 5 | + C \end{aligned}

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5.

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

Solution.

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

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6.

$\int t \sin (t) d t$

Solution.

Use integration by parts with

\begin{aligned} u & = t & d v & = \sin (t) d t \\ d u & = d t & v & = - \cos (t) \end{aligned}

Then

\begin{aligned} \int t \sin (t) d t & = t \cdot (- \cos (t)) - \int - \cos (t) d t \\ = - t \cos (t) + \int \cos (t) d t \\ = - t \cos (t) + \sin (t) + C \end{aligned}

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7.

$\int 5 x^{2} \cos (2 x^{3}) d x$

Solution.

Use substitution with $u = 2 x^{3}$ and $d u = 6 x^{2} d x,$ so $\frac{1}{6} d u = x^{2} d x :$

\begin{aligned} \int 5 x^{2} \cos (2 x^{3}) d x & = \int 5 \frac{1}{6} \cos (u) d u \\ = \frac{5}{6} \sin (u) + C \\ = \frac{5}{6} \sin (2 x^{3}) + C \end{aligned}

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8.

$\int \frac{1}{\sqrt{z + 8}} d z$

Solution.

Use substitution with $u = z + 8$ and $d u = d z :$

\begin{aligned} \int \frac{1}{\sqrt{z + 8}} d z & = \int \frac{1}{u} d u \\ = \int u^{- 1 / 2} d u \\ = \frac{u^{1 / 2}}{1 / 2} + C \\ = 2 (z + 8)^{1 / 2} + C \end{aligned}