4.9E: Exercises for Section 4.9

In exercises 1 - 20, find the antiderivative \(F(x)\) of each function \(f(x).\)

1) \(f(x)=\dfrac{1}{x^2}+x\)

2) \(f(x)=e^x−3x^2+\sin x\)

Answer

\(F(x)=e^x−x^3−\cos x+C\)

3) \(f(x)=e^x+3x−x^2\)

4) \(f(x)=x−1+4\sin(2x)\)

Answer

\(F(x)=\dfrac{x^2}{2}−x−2\cos(2x)+C\)

5) \(f(x)=5x^4+4x^5\)

6) \(f(x)=x+12x^2\)

Answer

\(F(x)=\frac{1}{2}x^2+4x^3+C\)

7) \(f(x)=\dfrac{1}{\sqrt{x}}\)

8) \(f(x)=\left(\sqrt{x}\right)^3\)

Answer

\(F(x)=\frac{2}{5}\left(\sqrt{x}\right)^5+C\)

9) \(f(x)=x^{1/3}+\big(2x\big)^{1/3}\)

10) \(f(x)=\dfrac{x^{1/3}}{x^{2/3}}\)

Answer

\(F(x)=\frac{3}{2}x^{2/3}+C\)

11) \(f(x)=2\sin(x)+\sin(2x)\)

12) \(f(x)=\sec^2 x +1\)

Answer

\(F(x)=x+\tan x+C\)

13) \(f(x)=\sin x\cos x\)

14) \(f(x)=\sin^2(x)\cos(x)\)

Answer

\(F(x)=\frac{1}{3}\sin^3(x)+C\)

15) \(f(x)=0\)

16) \(f(x)=\frac{1}{2}\csc^2 x+\dfrac{1}{x^2}\)

Answer

\(F(x)=−\frac{1}{2}\cot x −\dfrac{1}{x}+C\)

17) \(f(x)=\csc x\cot x+3x\)

18) \(f(x)=4\csc x\cot x−\sec x\tan x\)

Answer

\(F(x)=−\sec x−4\csc x+C\)

19) \(f(x)=8(\sec x)\big(\sec x−4\tan x\big)\)

20) \(f(x)=\frac{1}{2}e^{−4x}+\sin x\)

Answer

\(F(x)=−\frac{1}{8}e^{−4x}−\cos x+C\)

For exercises 21 - 29, evaluate the integral.

21) \(\displaystyle ∫(−1)\,dx\)

22) \(\displaystyle ∫\sin x\,dx\)

Answer

\(\displaystyle ∫\sin x\,dx = −\cos x+C\)

23) \(\displaystyle ∫\big(4x+\sqrt{x}\big)\,dx\)

24) \(\displaystyle ∫\frac{3x^2+2}{x^2}\,dx\)

Answer

\(\displaystyle ∫\frac{3x^2+2}{x^2}\,dx=3x−\frac{2}{x}+C\)

25) \(\displaystyle ∫\big(\sec x\tan x+4x\big)\,dx\)

26) \(\displaystyle ∫\big(4\sqrt{x}+\sqrt[4]{x}\big)\,dx\)

Answer

\(\displaystyle ∫\big(4\sqrt{x}+\sqrt[4]{x}\big)\,dx=\frac{8}{3}x^{3/2}+\frac{4}{5}x^{5/4}+C\)

27) \(\displaystyle ∫\left(x^{−1/3}−x^{2/3}\right)\,dx\)

28) \(\displaystyle ∫\frac{14x^3+2x+1}{x^3}\,dx\)

Answer

\(\displaystyle ∫\frac{14x^3+2x+1}{x^3}\,dx=14x−\frac{2}{x}−\frac{1}{2x^2}+C\)

29) \(\displaystyle ∫\big(e^x+e^{−x}\big)\,dx\)

In exercises 30 - 34, solve the initial value problem.

30) \(f′(x)=x^{−3},\quad f(1)=1\)

Answer

\(f(x)=−\dfrac{1}{2x^2}+\dfrac{3}{2}\)

31) \(f′(x)=\sqrt{x}+x^2,\quad f(0)=2\)

32) \(f′(x)=\cos x+\sec^2(x),\quad f(\frac{π}{4})=2+\frac{\sqrt{2}}{2}\)

Answer

\(f(x)=\sin x+\tan x+1\)

33) \(f′(x)=x^3−8x^2+16x+1,\quad f(0)=0\)

34) \(f′(x)=\dfrac{2}{x^2}−\dfrac{x^2}{2},\quad f(1)=0\)

Answer

\(f(x)=−\frac{1}{6}x^3−\dfrac{2}{x}+\dfrac{13}{6}\)

In exercises 35 - 39, find two possible functions \(f\) given the second- or third-order derivatives

35) \(f''(x)=x^2+2\)

36) \(f''(x)=e^{−x}\)

Answer

Answers may vary; one possible answer is \(f(x)=e^{−x}\)

37) \(f''(x)=1+x\)

38) \(f'''(x)=\cos x\)

Answer

Answers may vary; one possible answer is \(f(x)=−\sin x\)

39) \(f'''(x)=8e^{−2x}−\sin x\)

40) A potato is shot up vertically from by a potato cannon with an initial velocity of \(86\) ft/s and an initial height of \(3\)ft. Assuming the acceleration due to gravity is \(-32\) ft/s\(^2\), find the equations which model the potato's velocity and position \(t\) seconds after being launched. When does the potato reach it's highest point?

Answer

 

\(v(t)= -32t+86\)

\(s(t)= -16t^2+86t+3\)

When did the potato reach it's highest point? Immediately after \(v(t) = 0\):

\(-32t+86 = 0 \quad \Rightarrow\quad t = \frac{43}{16} \approx 2.69\text{s}.\)

41) A rock is launched upward from the edge of a \(20\) meter tall cliff. Three seconds after being launched, the rock has an upward velocity of \(19.6\)m/s. Assuming the acceleration due to gravity is \(-9.8\) m/s\(^2\), find the equations which model the rock's velocity and position \(t\) seconds after being launched. At what time does the rock hit the ground? If necessary, round your answer to the nearest hundredth of a second.

Hint

\(v(3) = 19.6\)

Answer

 

\(v(t)= -9.8t+49\)

\(s(t)= -4.9t^2+49t+20\)

The time at which the rock hits the ground is at \(t \approx 10.39\) seconds.

42) A car is being driven at a rate of \(40\) mph when the brakes are applied. The car decelerates at a constant rate of \(10\, \text{ft/sec}^2\). How long before the car stops?

Answer

\(5.867\) sec

43) In the preceding problem, calculate how far the car travels in the time it takes to stop.

44) You are merging onto the freeway, accelerating at a constant rate of \(12\, \text{ft/sec}^2\). How long does it take you to reach merging speed at \(60\) mph?

Answer

\(7.333\) sec

45) Based on the previous problem, how far does the car travel to reach merging speed?

46) A car company wants to ensure its newest model can stop in \(8\) sec when traveling at \(75\) mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

Answer

\(13.75\, \text{ft/sec}^2\)

47) A car company wants to ensure its newest model can stop in less than \(450\) ft when traveling at \(60\) mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

In exercises 48 - 53, find the antiderivative of the function, assuming \(F(0)=0.\)

48) [T] \(\quad f(x)=x^2+2\)

Answer

\(F(x)=\frac{1}{3}x^3+2x\)

49) [T] \(\quad f(x)=4x−\sqrt{x}\)

50) [T] \(\quad f(x)=\sin x+2x\)

Answer

\(F(x)=x^2−\cos x+1\)

51) [T] \(\quad f(x)=e^x\)

52) [T] \(\quad f(x)=\dfrac{1}{(x+1)^2}\)

Answer

\(F(x)=−\dfrac{1}{x+1}+1\)

53) [T] \(\quad f(x)=e^{−2x}+3x^2\)

In exercises 54 - 57, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false.

54) If \(f(x)\) is the antiderivative of \(v(x)\), then \(2f(x)\) is the antiderivative of \(2v(x).\)

Answer

True

55) If \(f(x)\) is the antiderivative of \(v(x)\), then \(f(2x)\) is the antiderivative of \(v(2x).\)

56) If \(f(x)\) is the antiderivative of \(v(x),\) then \(f(x)+1\) is the antiderivative of \(v(x)+1.\)

Answer

False

57) If \(f(x)\) is the antiderivative of \(v(x)\), then \((f(x))^2\) is the antiderivative of \((v(x))^2.\)