# Virtual Math Learning Center

## Practice Problems for Module 1

Covers Sections 5.5 and 6.1

### Exercises

Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.

1. Apply $$u$$-substitution to evaluate the integral: ​$$\displaystyle \int \! xe^{7x^2} \, dx$$

​$$\displaystyle \int \! xe^{7x^2} \, dx = \dfrac{1}{14}e^{7x^2} + C$$

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2. Apply $$u$$-substitution to evaluate the integral: ​$$\displaystyle \int \! \frac{x^2+2}{x^3+6x}\, dx$$

$$\displaystyle \int \! \frac{x^2+2}{x^3+6x}\, dx = \frac{1}{3}\ln\left|x^3+6x\right|+C$$

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3. Apply $$u$$-substitution to evaluate the integral: ​$$\displaystyle \int \! \frac{5+4x}{x^2+1} \, dx$$

$$\displaystyle \int \! \frac{5+4x}{x^2+1} \, dx = 5\arctan (x)+2\ln \left| x^2+1\right| + C$$

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4. Apply $$u$$-substitution to evaluate the integral: $$\displaystyle \int \! x\sqrt{x+5} \, dx$$

$$\displaystyle \int \! x\sqrt{x+5} \, dx = \frac{10}{3} (x+5)^{3/2} - \frac{2}{5}(x+5)^{5/2} + C$$

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5. Apply $$u$$-substitution to evaluate the integral: $$\displaystyle \int_0^1 \! 5x^2\sin\left(4x^3-7\right) \, dx$$

$$\displaystyle \int_0^1 \! 5x^2\sin\left(4x^3-7\right) \, dx = -\frac{5}{12}\cos(-3) + \frac{5}{12} \cos (-7)$$

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6. Apply $$u$$-substitution to evaluate the integral: $$\displaystyle \int_1^2 \! x^3\left(1-x^2\right)^5 \, dx$$

$$\displaystyle \int_1^2 \! x^3\left(1-x^2\right)^5 \, dx = \frac{1}{2}\left[ \frac{(-3)^7}{7} - \frac{(-3)^6}{6}\right]$$

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7. Sketch the region enclosed by the curves and find its area.
$y=2x^2+5 \qquad \textrm{ and } \qquad y=5x^2-7$

The area is 32.

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8. Sketch the region enclosed by the curves. Set up the integral with respect to both $$x$$ and $$y$$ that would give the area of the region.
$y=x+3 \qquad \textrm{ and } \qquad y=\sqrt{2x+6}$

The integral giving the area is $\displaystyle \int_{-3}^{-1} \! \sqrt{2x+6} - x -3 \, dx$ or $\displaystyle \int_0^2 \! y- \frac{1}{2}y^2 \, dy$

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9. Find the area of the region in the first quadrant that is bounded by the curves:
$xy=12, \qquad 3y=x, \qquad \textrm{ and } \qquad 3y=4x$

Video Errata: You must set $$x/3 = 4x/3$$ to get the point (0, 0). The video said setting these equal would not give any new points, which is not true. There was also a minor typo when recording, and the presenter had to manually write $$3$$ in $$3y = 4x$$.

The area is $$12 \ln (2)$$.

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10. Sketch the region bounded by the curve $$y = e^{x/2}$$, the tangent line to this curve at $$x$$ = 3, the $$x$$-axis and the $$y$$-axis. Set up the integral(s) representing the area of this region.

Video Errata: $$e^{3x/2}$$ was written in some spots where $$e^{x/2}$$ should have been written:
when finding the point of intersection $$(0, 1)$$ and when writing the final integrals.

The area is represented by $\displaystyle \int_0^1 \! e^{x/2} \, dx + \int_1^3 \! e^{x/2}-\frac{1}{2}e^{3/2}(x-1)\, dx$

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