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: \(\displaystyle \int \! xe^{7x^2} \, dx\)
\(\displaystyle \int \! xe^{7x^2} \, dx = \dfrac{1}{14}e^{7x^2} + C\)
2. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int \! \frac{x^2+2}{x^3+6x}\, dx\)
\(\displaystyle \int \! \frac{x^2+2}{x^3+6x}\, dx = \frac{1}{3}\ln\left|x^3+6x\right|+C\)
3. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int \! \frac{5+4x}{x^2+1} \, dx\)
\(\displaystyle \int \! \frac{5+4x}{x^2+1} \, dx = 5\arctan (x)+2\ln \left| x^2+1\right| + C\)
4. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int \! x\sqrt{x+5} \, dx\)
\(\displaystyle \int \! x\sqrt{x+5} \, dx = \frac{10}{3} (x+5)^{3/2} - \frac{2}{5}(x+5)^{5/2} + C\)
5. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int_0^1 \! 5x^2\sin\left(4x^3-7\right) \, dx\)
\(\displaystyle \int_0^1 \! 5x^2\sin\left(4x^3-7\right) \, dx = -\frac{5}{12}\cos(-3) + \frac{5}{12} \cos (-7)\)
6. Apply \(u\)-substitution to evaluate the integral: \(\displaystyle \int_1^2 \! x^3\left(1-x^2\right)^5 \, dx\)
\(\displaystyle \int_1^2 \! x^3\left(1-x^2\right)^5 \, dx = \frac{1}{2}\left[ \frac{(-3)^7}{7} - \frac{(-3)^6}{6}\right]\)
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.
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 \[\displaystyle \int_0^2 \! y- \frac{1}{2}y^2 \, dy\]
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 \(12 \ln (2)\).
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 \[\displaystyle \int_0^1 \! e^{x/2} \, dx + \int_1^3 \! e^{x/2}-\frac{1}{2}e^{3/2}(x-1)\, dx\]