Practice Problems

Work through the odd-numbered problems 1-29. Once you have completed the problem set, check your answers.

Practice Problems

Answers

1. \int_{0}^{4} 2+3 x d x

3. \int_{2}^{5} x^{3} d x


5. \int_{1}^{5} x^{3} d x

7. \int_{0.5}^{2} x \sin (x) d x

9. \int_{1}^{3} \ln (x) d x


11. \int_{1}^{3} 2 x d x=8

13.  \int_{-1}^{0}|x| \mathrm{dx}=1 / 2

15. \begin{aligned} &\int_{0}^{4} 3-\frac{x}{2} d x=8 \\ \end{aligned}


17. (a) 3 (b) –1 (c) 6 (d) 8 (e) 7


19.
(a) see the graph
(b) 24 feet.
(c) 24 feet from the starting point.



21. meters

23. feet3 = cubic feet

25. gram.meters

27. feet/second = feet per second


29. \Delta x=\frac{2-0}{n}=\frac{2}{n} \cdot m_{i}=\frac{2}{n}(i-1) \text { and } M_{i}=\frac{2}{n} i \text { so } f\left(m_{i}\right)=\left\{\frac{2}{n}(i-1)\right\}^{3} \text { and } f\left(M_{i}\right)=\left\{\frac{2}{n} i\right\}^{3}

a. \mathrm{LS}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{m}_{\mathrm{i}}\right) \Delta \mathrm{x}=\sum_{\mathrm{i}=1}^{\mathrm{n}}\left\{\frac{2}{\mathrm{n}}^{(\mathrm{i}-1)}\right\}^{3} \frac{2}{\mathrm{n}}=\frac{2}{\mathrm{n}} \frac{8}{\mathrm{n}^{3}}\left\{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}^{3}-3 \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}^{2}+3 \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}-\sum_{\mathrm{i}=1}^{\mathrm{n}} 1\right\}

=\frac{16}{n^{4}}\left\{\left(\frac{1}{4} n^{4}+\frac{1}{2} n^{3}+\frac{3}{12} n^{2}\right)-3\left(\frac{1}{3} n^{3}+\frac{1}{2} n^{2}+\frac{2}{12} n\right)+3\left(\frac{1}{2} n^{2}+\frac{1}{2} n\right)-n\right\}

=\frac{16}{n^{4}}\left\{\frac{1}{4} n^{4}-\frac{1}{2} n^{3}+\frac{1}{4} n^{2}\right\}=4-\frac{8}{n}+\frac{4}{n^{2}} \longrightarrow 4.

b. \mathrm{US}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{M}_{\mathrm{i}}\right) \Delta \mathrm{x}=\sum_{\mathrm{i}=1}^{\mathrm{n}}\left\{\frac{2}{\mathrm{n}} \mathrm{i}\right\}^{3} \frac{2}{\mathrm{n}}=\frac{2}{\mathrm{n}} \frac{8}{\mathrm{n}^{3}}\left\{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}^{3}\right\}=\frac{16}{\mathrm{n}}\left\{\frac{1}{4} \mathrm{n}^{4}+\frac{1}{2} \mathrm{n}^{3}+\frac{3}{12} \mathrm{n}^{2}\right\}

=4+\frac{8}{n}+\frac{4}{n^{2}} \longrightarrow 4.