Practice Problems

Site: Saylor Academy
Course: MA005: Calculus I
Book: Practice Problems
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Date: Saturday, September 7, 2024, 7:46 PM

Description

Work through the odd-numbered problems 1-41. Once you have completed the problem set, check your answers.

Table of contents

Practice Problems

In problems 1 – 3, use the values in Table 1 to estimate the areas.

x f(x) g(x) h(x)
0 5 2 5
1 6 1 6
2 6 2 8
3 4 2 6
4 3 3 5
5 2 4 4
6 2 5 2


1. Estimate the area between f and g for 1 \leq \mathrm{x} \leq 4.

3. Estimate the area between f and h for 0 \leq x \leq 4.

5. Estimate the area of the island in Fig. 13.



In problems 7 – 17, sketch the graph of each function and find the area between the graphs of f and g for x in the given interval.

7. \mathrm{f}(x)=x^{2}+3, \mathrm{~g}(x)=1 and -1 \leq x \leq 2.

9. \mathrm{f}(x)=x^{2}, \mathrm{~g}(x)=x and 0 \leq x \leq 2.

11. \mathrm{f}(x)=\frac{1}{x}, \mathrm{~g}(x)=x and 1 \leq x \leq \mathrm{e}.

13. \mathrm{f}(x)=x+1, \mathrm{~g}(x)=\cos (x) and 0 \leq x \leq \pi / 4.

15. f(x)=\mathrm{e}^{x}, \mathrm{~g}(x)=x and 0 \leq x \leq 2.

17. f(x)=3, g(x)=\sqrt{1-x^{2}} and 0 \leq x \leq 1.


In problems 19 – 21, use the values in Table 1 to estimate the average values.

19. Estimate the average value of f on the interval [0.5, 4.5].

21. Estimate the average value of f on the interval [1.5, 3.5].


In problems 23 – 31, find the average value of f on the given interval.

23. f(x) in Fig. 14 for 0 \leq x \leq 2.

25. f(x) in Fig. 14 for1 \leq x \leq 6.

27. \mathrm{f}(x)=2 x+1 for 0 \leq x \leq 4.

29. \mathrm{f}(x)=x^{2} for 1 \leq x \leq 3.

31. \begin{aligned}
    &\mathrm{f}(x)=\sin (x) \text { for } 0 \leq x \leq \pi \\ \end{aligned}.


33. Calculate the average value of \mathrm{f}(x)=\sqrt{x} on the interval [0, C] for C = 1, 9, 81, 100. What is the pattern?

35. Fig. 15 shows the number of telephone calls per minute at a large company. (a) Estimate the average number of calls per minute from 8 am to 5 pm. (b) From 9 am to 1 pm.


37. (a) How much work is done lifting a 20 pound bucket from the ground to the top of a 30 foot building with a cable which weighs 3 pounds per foot? (b) How much work is done lifting the same bucket from the ground to a height of 15 feet with the same cable?

39. (a) How much work is done lifting a 10 pound calculus book from the ground to the top of a 30 foot building with a cable which weighs 2 pounds per foot? (b) From the ground to a height of 10 feet? (c) From a height of 10 feet to a height of 20 feet?

41. How much work is done lifting an 60 pound injured child to the top of a 15 foot hole using a stretcher weighing 10 pounds and a cable which weighs 2 pound per foot?


Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.8-First-Applications-of-Definite-Integrals.pdf
Creative Commons License This work is licensed under a Creative Commons Attribution 3.0 License.

Answers

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 }