### Exercises

**Directions:** You should try to solve each problem first, and then click "Reveal Answer" to check your answer.
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1. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int \! xe^{7x^2} \, dx\)

Reveal Answer

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

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

Reveal Answer

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

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

Reveal Answer

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

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

Reveal Answer

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

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

Reveal Answer

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

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

Reveal Answer

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

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7. Sketch the region enclosed by the curves and find its area.

\[y=2x^2+5 \qquad \textrm{ and } \qquad y=5x^2-7\]

Reveal Answer

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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.

\[y=x+3 \qquad \textrm{ and } \qquad y=\sqrt{2x+6}\]

Reveal Answer

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 \(3y = 4x\).

Reveal Answer

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.

Reveal Answer

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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