Practice Problems

1. between 11 (using left endpoints of intervals) and 6 (using right endpoints)

3. between 4 (using left endpoints of intervals) and 6 (using right endpoints)

5. Using left endpoint widths: $(0)(40)+(70)(40)+(55)(40)+(90)(40)+(130)(40)+(115)(40)=18,400 \mathrm{ft}^{2}$

Right endpoint widths $(70,55, \ldots)$ and average widths $(70 / 2,125 / 2, \ldots)$ give the same result $18,400 \mathrm{ft}^{2}$. All of these are reasonable methods for estimating the area of the island.

7. $9$

9. $1$

11. $\frac{1}{2} \cdot \mathrm{e}^{2}-\frac{3}{2}$

13. $\frac{1}{32} \pi^{2}+\frac{1}{4} \pi-\frac{\sqrt{2}}{2}$

15. $e^{2}-3$

17. $3-\frac{\pi}{4}$

19. Estimate using midpoints of unit intervals: $\frac{1}{4}\{\mathrm{f}(1)(1)+\mathrm{f}(2)(1)+\mathrm{f}(3)(1)+\mathrm{f}(4)(1)\}=\frac{19}{4}$. About $\frac{19}{4}$.

21. Estimate using midpoints of unit intervals: $\frac{1}{2}\{\mathrm{f}(2)(1)+\mathrm{f}(3)(1)\}=5$. About 5.

23. $\text { average } \approx 1$

25. $\text { average } \approx \frac{11}{5}$

27. $\text { average }=5$

29. $\text { average }=\frac{13}{3}$

31. $\text { average }=\frac{2}{\pi}$

33.

(a) $\mathrm{C}=1: \text { average }=\frac{2}{3}$

(b) $\mathrm{C}=9: \text { average }=2$

(c) $\mathrm{C}=81: \text { average }=6$

(d) $\mathrm{C}=100: \text { average }=\frac{20}{3}$

In general, $\text { average }=\frac{2}{3} \sqrt{\mathrm{C}}$.

35.

(a) Graphically, $\text { average } \approx 3000 \cdot 1000 \frac{\text { calls }}{\text { hour }}=\frac{3000000}{60} \frac{\text { calls }}{\mathrm{min}} \approx 50,000 \frac{\text { calls }}{\mathrm{min}}$

(b) About $58,333 \frac{\text { calls }}{\mathrm{min}}$

37.

(a) Similar to Example 5: $\text { work }=1,950 \text { foot-pounds }$

(b) $\text { work }=1,312.5 \text { foot-pounds }$

39.

(a) $\text { work } = 1,200 foot–pounds$

(b) $\text { work }=600 \text { foot-pounds }$

(c) $\text { work }=400 \text { foot-pounds }$

41. $\text { work }=1,275 \text { foot-pounds }$