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Practice Problems for Module 1

Sections 5.5 and 6.1

Directions. The following are review problems. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link to the full video page to find related videos. 
  1. Apply \(u\)-substitution to evaluate the following integrals.
    1. ​\(\displaystyle \int \! xe^{7x^2} \, dx\)

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

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      WIR 20B M152 V1

    2. \(\displaystyle \int \! \frac{x^2+2}{x^3+6x}\, dx\)

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

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      WIR 20B M152 V2

    3. \(\displaystyle \int \! \frac{5+4x}{x^2+1} \, dx\)

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

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      WIR 20B M152 V3

    4. \(\displaystyle \int \! x\sqrt{x+5} \, dx\)

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

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      WIR 20B M152 V4

    5. \(\displaystyle \int_0^1 \! 5x^2\sin\left(4x^3-7\right) \, dx\)

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

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      WIR 20B M152 V5

    6. \(\displaystyle \int_1^2 \! x^3\left(1-x^2\right)^5 \, dx\)

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

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      WIR 20B M152 V6

  2. 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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    WIR 20B M152 V7

  3. 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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    WIR 20B M152 V8

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

    The area is \(12 \ln (2)\). 
    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 \(3y = 4x\).



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    WIR 20B M152 V9

  5. 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.

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



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    WIR 20B M152 V10