Unit #10 - Graphs of Antiderivatives, Substitution Integrals

Some problems and solutions selected or adapted from Hughes-Hallett Calculus.

Graphs of Antiderivatives

In Questions 1 to 4, sketch two functions F such thatF ′ = f . In one case, let F (0) = 0, and in the other letF (0) = 1.

1.

Solution:

2.

Solution:

3.

Solution:

4.

Solution:

5. Using the graph below, and the fact that P = 2when t = 0 to find values of P when t = 1, 2,3, 4 and 5.

0

1

−1

1 2 3 4 5

t

dP

dt

SincedP

dtis negative for t < 3 and positive for t > 3,

we know that P is decreasing for t < 3 and increasingfor t > 3. Between each two integer values, the mag-nitude of the change is equal to the area between the

graphdP

dtand the t-axis. For example, between t = 0

and t = 1, we see that the change in P is -1. SinceP = 2 at t = 0, we must have P = 1 at t = 1. Theother values are found similarly, and are shown in thetable below.

t 1 2 3 4 5P 1 0 -1/2 0 1

1

6. Given the values of the derivative f ′(x) in thetable and that f(0) = 100, use the TRAP ruleto estimate f(x) for x = 2, 4, 6.

x 0 2 4 6f ′(x) 10 18 23 25

By the Fundamental Theorem of Calculus, we knowthat

f(2)− f(0) =

∫ 2

0

f ′(x) dx.

Moving the f(0) to the right side and using the TRAPrule on this interval to estimate the integral,

f(2) = f(0) +

(10 + 18

2

)︸ ︷︷ ︸f estimate

2︸︷︷︸∆x

= 100 + 28 = 128.

Similarly, we can estimate f(4) based on f(2):

f(4) = f(2) +

(18 + 23

2

)︸ ︷︷ ︸f estimate

2︸︷︷︸∆x

= 128 + 41 = 169.

Finally, we can estimate f(6) based on f(4):

f(6) = f(4) +

(23 + 25

2

)︸ ︷︷ ︸f estimate

2︸︷︷︸∆x

= 169 + 48 = 217.

The final computed values are shown in summary in

this table:x 0 2 4 6≈ f(x) 100 128 169 217

In Questions 7 to 10, sketch two functions Fsuch that F ′ = f . In one case, let F (0) = 0,and in the other let F (0) = 1.Mark the points x1, x2, and x3 on the x-axis ofyour graph. Identify local maxima, minima andinflection points of F (x).

7.

F (x) has critical points at x1 and x3. Because of thesign changes in f (or F ′), x1 is a local max, and x3 isa local min. The point x2 is a point of inflection forF (x).

8.

F (x) has critical points at x1 and x3. Because of thesign changes in f (or F ′), x1 is a local max, and x3 isa local min. The point x2 is a point of inflection forF (x).

9.

F (x) has critical points at x1 and x3. Because of thesign changes in f (or F ′), x1 is a local min, and x3 isa local max. The point x2 is a point of inflection forF (x).

10.

2

F (x) has critical points at x1 and x3. However, be-cause f = F ′ is always positive, the function is alwaysincreasing, so x1 and x3 are neither mins nor maxes.In fact, all three points (x1, x2 and x3) are points ofinflection.

11. A particle moves back and forth along the x-axis. The graph below approximates the veloc-ity of the particle as a function of time. Positivevelocities represent movement to the right andnegative velocities represent movement to theleft. The particle starts at the point x = 5.Graph the distance of the particle from the ori-gin, with distance measured in kilometers andtime in hours.

Between t = 0 and t = 1, the particle moves at 10km/hr for 1 hour. Since it starts at x = 5, the particleis at x = 15 when t = 1. The graph of distance isa straight line between t = 0 and t = 1 because thevelocity is constant then. Between t = 1 and t = 2,the particle moves 10 km to the left, ending at x = 5.Between t = 2 and t = 3, it moves 10 km to the rightagain.

Here is a graph of the resulting position of the particle.

As an aside, note that the original velocity graph isnot entirely realistic as it suggests the particle reversesdirection instantaneously at the end of each hour. Inpractice this means the reversal of direction occurs overa time interval that is short in comparison to an hour.

12. Assume f ′ is given by the graph shown below.Suppose f is continuous and that f(3) = 0.

(a) Sketch a graph of f .

(b) Find f(0) and f(7).

(c) Find

∫ 7

0

f(x) dx in two different ways.

(a) Starting at x = 3, we are given that f(3) = 0.Moving to the left on the interval 2 < x < 3, wehave f ′(x) = −1. Since

f(3)− f(2) =

∫ 3

2

f ′(x) dx = −1,

so f(2) = f(3)− (−1) = 1.

On the interval 0 < x < 2, we have f ′(x) = 1, sof(0) = f(2) + 1(−2) = −1.

Moving to the right from x = 3, we know thatf ′(x) = 2 on 3 < x < 4. So

f(4) = f(3) + 2 = 2.

On the interval 4 < x < 6, f ′(x) = −2 so

f(6) = f(4) + 2(−2) = −2.

On the interval 6 < x < 7, we have f ′(x) = 1, so

f(7) = f(6) + 1 = −2 + 1 = −1.

3

(b) In part (a) we found that f(0) = −1 and f(7) =−1.

(c) The integral

∫ 7

0

f ′(x) dx is given by the sum

∫ 7

0

f ′(x) dx =(1)(2) + (−1)(1) + (2)(1)

+ (−2)(2) + (1)(1)

= 0

Alternatively, knowing f(7) and f(0) and using theFundamental Theorem of Calculus, we have∫ 7

0

f ′(x) dx = f(7)− f(0) = −1− (−1) = 0.

13. Use the graph of F ′(x) below and the fact thatF (2) = 3 to sketch the graph of F (x). Labelthe values of at least four points.

We can start by finding four points on the graph ofF(x). The first one is given: F (2) = 3. By the Funda-mental Theorem of Calculus,

F (6) = F (2) +

∫ 6

2

F ′(x) dx.

The value of this integral is -7 (the area is 7, but thegraph lies below the x-axis), so F (6) = 3−7 = −4. Sim-ilarly, F (0) = F (2)− 2 = 1, and F (8) = F (6) + 4 = 0.We sketch a graph of F (x) by connecting these points,as shown below.

14. Using the graph of g(t) below, sketch a graph ofan antiderivative G(t) of g(t) satisfying G(0) =5. Label each critical point of G(t) with its co-ordinates.

The critical points are at (0; 5), (2; 21), (4; 13), and(5; 15). A graph is given below.

15. Using the graph of g′ shown below, and the factthat g(0) = 50, sketch the graph of g(x). Givethe coordinates of all critical points and inflec-tion points of g.

Looking at the graph of g′, we see that the criticalpoints of g occur at x = 15 and x = 40, since g′(x) = 0at these values.

Inflection points of g occur when x = 10 and x = 20,because g′(x) has a local maximum or minimum atthese values.

4

Knowing these four key points, we build up points onthe graph of g(x).

We start at x = 0, where g(0) = 50 (given). Since g′

is negative on the interval [0, 10], the value of g(x) isdecreasing there. At x = 10 we have

g(10) = g(0) +

∫ 10

0

g′(x) dx

= 50− (area of shaded trapezoid T1)

= 50−(

10 + 20

2

)· 10

= −100.

Similarly,

g(15) = g(10) +

∫ 15

10

g′(x) dx

= −100− (area of triangle T2)

= −100− 1

2(5)(20) = −150.

Continuing,

g(20) = g(15) +

∫ 20

15g′(x) dx

= −150 +1

2(5)(10)

= −125,

and

g(40) = g(20) +

∫ 40

20g′(x) dx

= −125 +1

2(20)(10) = −25.

This gives us all the points we need to sketch g(x):

16. The Quabbin Reservoir in the western part ofMassachusetts provides most of Boston’s water.The graph below represents the flow of water inand out of the Quabbin Reservoir throughout2007.

(a) Sketch a graph of the quantity of water inthe reservoir, as a function of time.

(b) When, in the course of 2007, was the quan-tity of water in the reservoir largest? Small-est? Mark and label these points on thegraph you drew in part (a).

(c) When was the quantity of water increas-ing most rapidly? Decreasing most rapidly?Mark and label these times on both graphs.

(d) By July 2008 the quantity of water in thereservoir was about the same as in January2007. Draw plausible graphs for the flowinto and the flow out of the reservoir forthe first half of 2008.

(a) For the first ∼2 months, the outflow is greater thanthe inflow, so the reservoir level is dropping. Afterthat, until July, the inflow is higher, so the reser-voir level is rising. From July until Jan 1994, theoutflow again becomes higher than the inflow, sothe reservoir level is dropping over that time pe-riod.

We can estimate the net change in water level bythe area between the inflow and outflow graphs.From that, we can get a sketch like that shownbelow.

5

(b) The critical points of the reservoir level functionare where the inflow and outflow lines cross:

• ∼ March: local minimum,

• ∼ July: local maximum,

From our sketch, the peak around July is a globalmaximum, but the global minimum will be at theend of the time interval shown, as much more wateris lost than gained between March and Jan 1994.

(c) The water level is increasing most rapidly when theinflow graph is above the outflow by the largestvertical amount. As we have seen in other exam-ples, this will occur when the inflow and outflowrates have the same slope. The same is true forwhen the reservoir level is decreasing most rapidly,except that the outflow graph must be above theinflow graph at that point.

The two points are shown on the diagram below,along with the slopes. The first point is the timewhen the reservoir level is increasing most quickly,while the second is when it is decreasing mostquickly.

(d) In order for the water to reach the same level s itwas in Jan ’93, the total amount of water which hasflowed out of the reservoir (area between graphswhere outflow is above inflow graph) reservoir mustequal the amount that flowed into the reservoir

(area between graphs where inflow is above out-flow). Since the reservoir is currently lower than itstarted, you would have to sketch a continuationof the graph that had

• The inflow rising above in outflow, so the reser-voir level would be increasing again, and

• the inflow and outflow balanced so that thetotal areas representing net inflow and outflowamounts for the reservoir are equal over theperiod from Jan 1993 to Jul 1994.

A reasonable sketch is shown below. We attemptedto make the area of A+C (total volume out of reser-voir) roughly equal to the area B+D (total volumeinto reservoir).

17. Consider the function

f(x) =

{−x+ 1 for 0 ≤ x ≤ 1x− 1 for 1 < x ≤ 2 .

The graph of this function is shown here:

(a) Find a continuous function F such thatF ′ = f and F (1) = 1. Hint: F (x) willbe a piecewise function.

(b) Use geometry to calculate the area underthe graph of f and above the x−axis be-tween x = 0 and x = 2 and show that itequals F (2)− F (0).

(c) Use parts (a) and (b) to verify the Funda-mental Theorem of Calculus for this exam-ple.

(a) This question is similar to some of the antideriva-tive sketching problems we have seen, but nowwhat we want the actual formula for the function.

We find F over the intervals 0 ≤ x ≤ 1 and1 ≤ x ≤ 2. For 0 ≤ x ≤ 1, we have f(x) = −x+ 1

6

and so taking antiderivatives gives F (x) of the form∫(−x+ 1)dx = −x

2

2+ x+ C.

Since we want F (1) = 1, we have 1 = − 12

2 + 1 +C,or C = 1

2 .

For the interval from x = 1 to 2, we have f(x) =x − 1 and so taking antiderivatives means F (x) isof the form∫

(x− 1)dx =x2

2− x+D (new constant, C already used)

Again, we want F (1) = 1, so 1 = 12

2 − 1 + D andD = 3

2 .

Thus,

F (x) =

−12x

2 + x+ 12 if 0 ≤ x ≤ 1

12x

2 − x+ 32 if 1 ≤ x ≤ 2.

(b) F (2)−F (0) = 32−

12 = 1. The area under the graph

of f is the sum of the areas of the two triangles,which is 1

2 + 12 = 1.

(c) The Fundamental Theorem of Calculus says∫ 2

0

f(x)dx = F (2)− F (0).

Since the value of the integral is just the area underthe curve, we have verified this in part (b).

Substitution Integrals

To practice computing integrals using substitutions, do as many of the problems from this section as you feel youneed. The problems trend from simple to the more complex.

Note: In the solutions to these problems, we always show the substitution used. On a test, if you can compute theantiderivative in your head, you do not need to go through all the steps shown here. They are included in thesesolutions as a learning and comprehension aid.

18.

∫tet

2

dt

Let w = t2, so dw = 2t dt or dt =1

2tdw∫

tet2

dt =

∫tew

(1

2tdw

)=

∫1

2ewdw =

1

2ew + C =

1

2et

2

+ C

Check:d

dt

1

2et

2

+ C =1

2(2t)et

2

= tet2

= original function in integral. "

19.

∫e3x dx

Let w = 3x, so dw = 3 dx or dx =dw

3∫e3x dx =

∫ewdw

3=

1

3ew + C =

1

3e3x + C

Check:d

dx

1

3e3x + C =

1

3e3x(3) = e3x

= original function in integral. "

NOTE: we will not show the differentiationcheck for any later questions, as the process is

always the same, and you should be comfort-able enough with the derivative rules to do thechecking independently. If you are uncertainabout any problem, contact your instructor.

20.

∫e−x dx

Let w = −x, so dw = −1 dx or dx = (−dw)∫e−x dx =

∫ew(−dw) = ew + C = −e−x + C

21.

∫25e−0.2t dt

Let w = −0.2t, so dw = −0.2 dt

or dt =dw

−0.2= −5dw∫

25e−0.2t dt =

∫25ew(−5dw) = −125ew + C

= −125e−0.2t + C

22.

∫t cos(t2) dt

7

Let w = t2, so dw = 2t dt

or dt =dw

2t∫t cos(t2) dt =

∫t cos(w)

dw

2t=

1

2sin(w) + C

=1

2sin(t2) + C

23.

∫sin(2x) dx

Let w = 2x, so dw = 2dx, or dx =1

2dw∫

sin(2x)dx =1

2

∫sin(w)dw = −1

2cos(w) + C

= −1

2cos(2x) + C

24.

∫sin(3− t) dt

Let w = 3− t, so dw = −1 dt or dt = −dw∫sin(3− t) dt =

∫sin(w) (−1)dw = −(− cos(w)) + C

= cos(3− t) + C

25.

∫xe−x

2

dx

Let w = −x2, so dw = −2x dx, or dx =−1

2xdw∫

xe−x2

dx =

∫xew−1

2xdw =

∫−1

2ew dw = −1

2ew + C

= −1

2e−x

2

+ C

26.

∫(r + 1)3 dr

Let w = (r + 1), so dw = dr∫(r + 1)3 dr =

∫w3 dw =

w4

4+ C =

(r + 1)4

4+ C

27.

∫y(y2 + 5)8 dy

Let w = y2 + 5, so dw = 2y dy, or dy =1

2ydw∫

y(y2 + 5)8 dy =

∫yw8 1

2ydw =

1

2

∫w8 dw =

1

2

w9

9+ C

=1

18(y2 + 5)9 + C

28.

∫t2(t3 − 3)10 dt

Let w = t3 − 3, so dw = 3t2 dt, or dt =1

3t2dw∫

t2(t3 − 3)10 dt =

∫t2w10 1

3t2dw =

1

3

∫w10 dw

=1

3

w11

11+ C =

1

33(t3 − 3)11 + C

29.

∫x2(1 + 2x3)2 dx

Let w = 1 + 2x3, so dw = 6x2 dx, or dx =1

6x2dw∫

x2(1 + 2x3)2 dx =

∫x2w2 1

6x2dw =

1

6

w3

3+ C

=1

18(1 + 2x3)3 + C

30.

∫x(x2 + 3)2 dx

Let w = x2 + 3, so dw = 2x dx, or dx =1

2xdw∫

x(x2 + 3)2 dx =

∫xw2 1

2xdw =

1

2

w3

3+ C

=1

6(x2 + 3)3 + C

31.

∫x(x2 − 4)7/2 dx

Let w = x2 − 4, so dw = 2x dx, or dx =1

2xdw∫

x(x2 − 4)7/2 dx =

∫xw7/2 1

2xdw =

1

2

w9/2

9/2+ C

=1

9(x2 − 4)9/2 + C

32.

∫y2(1 + y)2 dy

Trick (substitution) question: substitution seems notto work well here, because both factors have y2 in them,so neither one is the derivative of the other. We’re bet-ter off expanding the (1 + y)2 factor and then integrat-ing each term separately:∫y2(1 + y)2dy =

∫y2(1 + 2y + y2)dy =

∫y2 + 2y3 + y4

=1

3y3 +

2

4y4 +

1

5y5 + C

8

33.

∫(2t− 7)73 dt

Let w = 2t− 7, so dw = 2 dt, or dt =1

2dw∫

(2t− 7)73 dt =

∫w73 1

2dw =

1

2

w74

74+ C

=1

148(2t− 7)74 + C

34.

∫1

y + 5dy

Let w = y + 5, so dw = dy, making∫dy

y + 5dy =

∫1

wdw = ln |w|+ C = ln |y + 5|+ C

35.

∫1√

4− xdx

Rewrite integral:∫(4− x)−1/2 dx

Let w = 4− x, so dw = −1 dx, or dx = (−1)dw∫(4− x)−1/2 dx =

∫w−1/2(−1)dw = −w

+1/2

1/2

= −2(4− x)1/2 + C

36.

∫(x2 + 3)2 dx

Another non-substitution integral: since there is no xterm outside the (x2 + 3), it is easier to expand thesquare in this case and integrate term by term.∫

(x2 + 3)2 dx =

∫x4 + 6x2 + 9 dx =

x5

5+

6x3

3+ 9x+ C

=x5

5+ 2x3 + 9x+ C

37.

∫x2ex

3+1 dx

Let w = x3 + 1, so dw = 3x2 dx, or dx =1

3x2dw∫

x2ex3+1 dx =

∫x2ew

1

3x2dw =

1

3ew + C

=1

3ex

3+1 + C

38.

∫sin(θ)(cos(θ) + 5)7 dθ

Let w = cos(θ) + 5, so dw = − sin(θ)dθ, making∫sin(θ)(cos(θ) + 5)7dθ = −

∫w7dw = −w

8

8+ C

= −1

8(cos(θ) + 5)8 + C

39.

∫ √cos(3t) sin(3t) dt

Let w = cos(3t) so dw = −3 sin(3t) dt,

or dt =−1

3 sin(3t)dw∫ √

cos(3t) sin(3t) dt =

∫w1/2 sin(3t)

−1

3 sin(3t)dw

=−1

3

w3/2

3/2+ C

=−2

9(cos(3t))3/2 + C

40.

∫sin6(θ) cos(θ) dθ

Let w = sin(θ) so dw = cos(θ) dθ,

or dθ =1

cos(θ)dw∫

(sin(θ))6 cos(θ) dθ =

∫w6 cos(θ)

1

cos(θ)dw =

w7

7+ C

=1

7sin7(θ) + C

41.

∫sin3(α) cos(α)dα

Let w = sin(α) so dw = cos(α) dα, or dα =1

cos(α)dw∫

(sin(α))3 cos(α) dα =

∫w3 cos(α)

1

cos(α)dw =

w4

4+ C

=1

4sin4(α) + C

42.

∫sin6(5θ) cos(5θ) dθ

Let w = sin(5θ) so dw = 5 cos(5θ) dθ,

or dθ =1

5 cos(5θ)dw∫

(sin(5θ))6 cos(5θ) dθ =

∫w6 cos(5θ)

1

5 cos(5θ)dw

=1

5

w7

7+ C =

1

35sin7(5θ) + C

9

43.

∫tan(2x) dx

Rewrite integral:∫sin(2x)

cos 2xdx

Let w = cos(2x), so dw = −2 sin(2x) dx,

or dx =−1

2 sin(2x)dw∫

sin(2x)

cos(2x)dx =

∫sin(2x)

w

−1

2 sin(2x)dw =

−1

2

∫w−1 dw

=−1

2ln(|w|) + C =

−1

2ln(| cos(2x)|) + C

44.

∫(ln z)2

zdz

Let w = ln(z), so dw =1

zdz, or dz = z dw∫

(ln z)2

zdx =

∫w2

z(z dw) =

w3

3+ C

=(ln z)3

3+ C

45.

∫et + 1

et + tdt

Let w = et + t, so dw = (et + 1) dt, or dt =1

et + 1dw∫

et + 1

et + tdt =

∫et + 1

w

1

et + 1dw =

∫w−1 dw

= ln(|w|) + C = ln(|et + t|) + C

46.

∫y

y2 + 4dy

Let w = y2 + 4, so dw = 2y dy, or dy =1

2ydw∫

y

y2 + 4dy =

∫y

w

1

2ydw =

1

2

∫w−1 dw

=1

2ln(|w|) + C =

1

2ln(|y2 + 4|) + C

Note that y2 + 4 is always positive, so we could removethe absolute values if we wished, as they are redundantin this case.

47.

∫cos(√x)√

xdx

Let w =√x, so dw =

1

2x−1/2 dx, or dx = 2

√x dw∫

cos(√x)√

xdx =

∫cos(w)√

x(2√x dw) = 2

∫cos(w) dw

= 2 sin(w) + C = 2 sin(√x) + C

48.

∫e√y

√ydy

Let w =√y, so dw =

1

2y−1/2 dy, or dy = 2

√y dw∫

e√y

√ydy =

∫ew√y

(2√y dw) = 2

∫ew dw

= 2ew + C = 2e√y + C

49.

∫1 + ex√x+ ex

dx

Let w = x+ ex, so dw = (1 + ex) dx, or dx =1

1 + exdw∫

1 + ex√x+ ex

dx =

∫1 + ex√

w

(1

1 + exdw

)=

∫w−1/2 dw

=w1/2

1/2+ C = 2

√x+ ex + C

50.

∫ex

2 + exdx

Let w = 2 + ex, so dw = ex dx, or dx =1

exdw∫

ex

2 + exdx =

∫ex

w

(1

exdw

)=

∫w−1 dw

= ln(|w|) + C = ln(|2 + ex|) + C

51.

∫x+ 1

x2 + 2x+ 19dx

Let w = x2 + 2x+ 19, so dw = (2x+ 2) dx,

or dx =1

2(x+ 1)dw

∫x+ 1

x2 + 2x+ 19dx =

∫x+ 1

w

(1

2(x+ 1)dw

)=

1

2

∫w−1 dw =

1

2ln(|w|) + C

=1

2ln(|x2 + 2x+ 19|) + C

10

52.

∫t

1 + 3t2dt

Let w = 1 + 3t2, so dw = 6t dt, or dt =1

6tdw∫

t

1 + 3t2dt =

∫t

w

(1

6tdw

)=

1

6

∫w−1 dw

=1

6ln(|w|) + C =

1

6ln(|1 + 3t2|) + C

In this case we can remove the absolute values because1 + 3t2 will always be positive, so the absolute valuesare redundant.

53.

∫ex − e−x

ex + e−xdx

Let w = ex + e−x, so dw = (ex − e−x) dx,

or dx =1

ex − e−xdw∫

ex − e−x

ex + e−xdx =

∫ex − e−x

w

(1

ex − e−xdw

)=

∫w−1 dw = ln(|w|) + C

= ln(|ex + e−x|) + C

In this case we can remove the absolute values becauseex + e−x will always be positive, so the absolute valuesare redundant.

54.

∫(t+ 1)2

t2dt

This question is probably more easily solved by expand-ing than by using substitution.∫

(t+ 1)2

t2dt =

∫t2 + 2t+ 1

t2dt =

∫1 +

2

t+

1

t2dt

= t+ 2 ln(|t|)− t−1 + C

= t+ 2 ln(|t|)− 1

t+ C

55.

∫x cos(x2)√

sin(x2)dx

Let w = sin(x2), so dw = 2x cos(x2) dx or

dx =dw

2x cos(x2)

∫x cos(x2)√

sin(x2)dx =

∫x cos(x2)√

w

(1

2x cos(x2)dw

)=

1

2

∫w−1/2 dw =

1

2

(w1/2

1/2

)+ C

=√w + C =

√sin(x2) + C

56.

∫ π

0

cos(x+ π) dx

∫ π

0

cos(x+ π) dx = sin(x+ π)∣∣∣π0

= sin(2π)− sin(π)

= 0− 0 = 0

57.

∫ 1/2

0

cos(πx) dx

∫ 1/2

0

cos(πx) dx =1

πsin(πx)

∣∣∣1/20

=1

π[sin(π/2)− sin(0)] =

1

π[1− 0] =

1

π

58.

∫ π/2

0

e− cos(θ) sin(θ) dθ

In this solution, we will use the substitution just to findthe antiderivative, but then we will switch back to theoriginal integral variable θ to evaluate the limits.

Let w = − cos(θ), so dw = sin(θ) dθ or dθ =dw

sin(θ)∫ π/2

0

e− cos(θ) sin(θ) dθ =

∫ θ=π/2

θ=0

ew sin(θ)

(dw

sin(θ)

)=

∫ θ=π/2

θ=0

ew dw = ew∣∣∣θ=π/2θ=0

= e− cos(θ)∣∣∣θ=π/2θ=0

= e0 − e−1 = 1− 1

e

59.

∫ 2

1

2xex2

dx

In this solution, we will convert the limits of integrationto the substitution variable.

Let w = x2, so dw = 2x dx or dx = (1/2x)dw

then also x = 1→ w = 12 = 1,

and x = 2→ w = 22 = 4.

∫ x=2

x=1

2xex2

dx =

∫ w=4

w=1

2xew(

1

2xdw

)=

∫ w=4

w=1

ew dw = ew∣∣∣w=4

w=1

= e4 − e1

11

60.

∫ 8

1

e3√x

3√x2

dx

In this solution, we will convert the limits of integrationto the substitution variable.

Let w = x1/3, so dw =1

3x−2/3 dx or dx = 3x2/3dw

then also x = 1→ w = (1)1/3 = 1,

and x = 8→ w = 81/3 = 2.

∫ x=8

x=1

e3√x

3√x2

dx =

∫ w=2

w=1

ew

x2/3

(3x2/3 dw

)=

∫ w=2

w=1

3ew dw = 3ew∣∣∣w=2

w=1= 3e2 − 3e1

61.

∫ e−2

−1

1

t+ 2dt

Let w = t+ 2, so dw = dt

then also t = −1→ w = (−1) + 2 = 1

and t = e− 2→ w = (e− 2) + 2 = e

∫ t=e−2

t=−1

1

t+ 2dt =

∫ w=e

w=1

1

wdw = ln(|w|)

∣∣∣w=e

w=1

= ln(e)− ln(1) = 1− 0 = 1

62.

∫ 4

1

cos√x√

xdx

Let w = x1/2 so dw =1

2x−1/2 dx or dx = 2x1/2 dw

then also x = 1→ w = (1)1/2 = 1

and x = 4→ w = (4)1/2 = 2

∫ x=4

x=1

cos√x√

xdx =

∫ w=2

w=1

cos(w)

x1/2(2x1/2 dw)

=

∫ w=2

w=1

2 cos(w) dw = 2 sin(w)∣∣∣w=2

w=1

= 2 sin(2)− 2 sin(1)

63.

∫ 2

0

x

(1 + x2)2dx

Let w = 1 + x2 so dw = 2x dx or dx =1

2xdw

then also x = 0→ w = 1 + 02 = 1

and x = 2→ w = 1 + 22 = 5

∫ x=2

x=0

x

(1 + x2)2dx =

∫ w=5

w=1

x

w2

(1

2xdw

)=

1

2

∫ w=5

w=1

1

w2dw =

1

2

(−1

w

) ∣∣∣w=5

w=1

=1

2

[−1

5− −1

1

]=

1

2

4

5=

2

5

64. If appropriate, evaluate the following integralsby substitution. If substitution is not appropri-ate, say so, and do not evaluate.

(a)

∫x sin(x2) dx

(b)

∫x2 sin(x) dx

(c)

∫x2

1 + x2dx

(d)

∫x

(1 + x2)2dx

(e)

∫x3ex

2

dx

(f)

∫sin(x)

2 + cos(x)dx

(a) This integral can be evaluated using integration bysubstitution. We use w = x2, dw = 2x dx.∫x sin(x2) dx =

1

2

∫sin(w) dw

= −1

2cos(w) + C =

−1

2cos(x2) + C

(b) This integral cannot be evaluated using a simpleintegration by substitution.

(c) This integral cannot be evaluated using a simpleintegration by substitution.

(d) This integral can be evaluated using integration bysubstitution. We use w = 1 + x2, dw = 2x dx.∫

x

(1 + x2)2dx =

1

2

∫1

w2dw =

1

2

(−1

w

)+ C

=−1

2(1 + x2)+ C.

(e) This integral cannot be evaluated using a simpleintegration by substitution.

(f) This integral can be evaluated using integrationby substitution. We use w = 2 + cosx, dw =

12

− sinx dx.∫sinx

2 + cosxdx = −

∫1

wdw = − ln |w|+ C

= − ln |2 + cosx|+ C :

65. Find the exact area under the graph of f(x) =

xex2

between x = 0 and x = 2.

Since f(x) is always positive, the area under the graphis equal to the integral of f(x) on that interval. Thuswe need to evaluate the definite integral∫ 2

0

xex2

dx.

This is done in Question 59, using the substitutionw = x2, the result being∫ 2

0

xex2

dx =1

2(e4 − 1).

66. Find the exact area under the graph of f(x) =1

x+ 1between x = 0 and x = 2.

Since f(x) = 1/(x+ 1) is positive on the interval x = 0to x = 2, the area under the graph and the integral areequal to one another.∫ 2

0

1

x+ 1dx = ln(x+ 1)

∣∣∣20

= ln 3− ln 1 = ln 3.

The area is ln 3 ≈ 1.0986.

67. Find

∫4x(x2 + 1) dx using two methods:

(a) Do the multiplication first, and then antid-ifferentiate.

(b) Use the substitution w = x2 + 1.

(c) Explain how the expressions from parts (a)and (b) are different. Are they both cor-rect?

(a)

∫4x(x2 +1) dx =

∫(4x3 +4x) dx = x4 +2x2 +C.

(b) If w = x2 + 1 then dw = 2x dx:∫4x(x2 + 1) dx =

∫2w dw = w2 + C = (x2 + 1)2 + C.

(c) The expressions from parts (a) and (b) look dif-ferent, but they are both correct. Note that theanswer from (b) can be expanded as

(x2 + 1)2 + C = x4 + 2x2 + 1 + C︸ ︷︷ ︸new const.

.

In other words, the expressions from parts (a) and(b) differ only by a constant, so they are both cor-rect antiderivatives.

68. (a) Find

∫sin θ cos θ dθ

(b) You probably solved part (a) by making thesubstitution w = sin θ or w = cos θ. (Ifnot, go back and do it that way.) Now find∫

sin θ cos θ dθ by making the other substi-

tution.

(c) There is yet another way of finding this inte-gral which involves the trigonometric iden-tities:

sin(2θ) = 2 sin θ cos θ

cos(2θ) = cos2 θ − sin2 θ.

Find

∫sin θ cos θ dθ using one of these

identities and then the substitution w = 2θ.

(d) You should now have three differentexpressions for the indefinite integral∫

sin θ cos θ dθ. Are they really different?

Are they all correct? Explain.

(a) We first try the substitution w = sin θ, so dw =cos θ dθ. Then∫

sin θ cos θ dθ =

∫w dw =

w2

2+ C =

sin2 θ

2+ C.

(b) If we instead try the substitution w = cos θ, dw =− sin θ dθ, we get∫

sin θ cos θ dθ = −∫w dw = −w

2

2+ C = −cos2 θ

2+ C.

(c) Once we note that sin(2θ) = 2 sin θ cos θ we canalso say ∫

sin θ cos θ dθ =1

2

∫sin(2θ) dθ

Substituting w = 2θ, dw = 2 dθ, the above equals

1

4

∫sinw dw = −cosw

4+ C = −cos 2θ

4+ C.

(d) All these answers are correct. Although they havedifferent forms, they differ from each other only interms of a constant, and thus they are all accept-able antiderivatives.

For example,

1− cos2 θ = sin2 θ

sosin2 θ

2︸ ︷︷ ︸Answer (a)

= −cos2 θ + 1

2= −cos2 θ

2︸ ︷︷ ︸Answer (b)

−1

2

13

Thus the first two expressions differ only by a con-stant.

Similarly, cos(2θ) = cos2 θ − sin2 θ = 2 cos2 θ − 1,so

−cos(2θ)

4︸ ︷︷ ︸Answer (c)

=− cos2 θ

2︸ ︷︷ ︸Answer (b)

+1

4

and thus the second and third expressions differonly by a constant. Of course, if the first two ex-pressions and the last two expressions differ only inthe constant C, then the first and last only differin the constant as well, so they are all equally validantiderivatives.

Substitution Integrals - Applications

69. Let f(t) be the rate of flow, in cubic meters perhour, of a flooding river at time t in hours. Givean integral for the total flow of the river:

(a) Over the 3-day period, 0 ≤ t ≤ 72 (since tis measured in hours).

(b) In terms of time T in days over the same3-day period.

(a) A time period of ∆t hours with flow rate of f(t)cubic meters per hour has a flow of f(t)∆t cu-bic meters. Summing the flows, we get total flow≈∑f(t)∆t, so

Total flow =

∫ 72

0

f(t) dt cubic meters

(b) Since 1 day is 24 hours, t = 24T . The con-stant 24 has units (hours/day), so 24T has units(hours/day) × day = hours. We can construct thenew integral using our earlier substitution process.

t = 24T

so 1 = 24dT

dtor dt = 24dT

For the limits of integration:T = t/24 so t = 0hours corresponds to T = 0/24 = 0 days, andt = 72 hours corresponds to T = 72/24 = 3 days.

Rewriting the integral from part (a) using this sub-stitution, we get

Total flow =

∫ T=3

T=0

f(24T ) (24dT )

= 24

∫ 3

0

f(24T ) dT cubic meters.

70. Oil is leaking out of a ruptured tanker at therate of r(t) = 50e−0.02t thousand liters perminute.

(a) At what rate, in liters per minute, is oilleaking out at t = 0? At t = 60?

(b) How many liters leak out during the firsthour?

(a) At time t = 0, the rate of oil leakage = r(0) = 50thousand liters/minute.At t = 60, rate = r(60) = 15.06 thousandliters/minute.

(b) To find the amount of oil leaked during the firsthour, we integrate the rate from t = 0 to t = 60:

Oil leaked =

∫ 60

0

50e−0.02tdt =

(−50

0.02e−0.02t

) ∣∣∣60

0

= −2500e−1.2 + 2500e0

≈ 1747 thousand liters.

71. If we assume that wind resistance is propor-tional to velocity, then the downward velocity,v, of a body of mass m falling vertically is givenby

v =mg

k(1− e−kt/m)

where g is the acceleration due to gravity andk is a constant. Find the height of the body, h,above the surface of the earth as a function oftime. Assume the body starts at height h0.

Since v = dh/dt, it follows that h(t) =∫v(t) dt and

h(0) = h0. Since

v(t) =mg

k

(1− e− k

m t)

=mg

k− mg

ke−

km t

we have

h(t) =

∫v(t) dt

=mg

k

∫dt− mg

k

∫e−k/mt dt.

14

The first integral is simplymg

kt+ C.

To evaluate the second integral, make the substitution

w = − kmt. Then dw = − k

mdt, so∫

e−kt/m dt =

∫ew(−mk

dw

)= −m

kew + C

= −mke−kt/m + C.

Thus

h(t) =

∫v dt

=mg

kt− mg

k

(−mke−kt/m

)+ C

=mg

kt+

m2g

k2e−kt/m + C.

Since h(0) = h0,

h0 =mg

k· 0 +

m2g

k2e0 + C;

C = h0 −m2g

k2.

Thus

h(t) =mg

kt+

m2g

k2e−kt/m − m2g

k2+ h0

h(t) =mg

kt− m2g

k2

(1− e−kt/m

)+ h0.

This formula gives the height of the object above thesurface of the earth as it falls.

72. The rate at which water is flowing into a tankis r(t) gallons/minute, with t in minutes.

(a) Write an expression approximating theamount of water entering the tank duringthe interval from time t to time t+∆t, where∆t is small.

(b) Write a Riemann sum approximating thetotal amount of water entering the tank be-tween t = 0 and t = 5. Then write an exactexpression for this amount.

(c) By how much has the amount of water inthe tank changed between t = 0 and t = 5if r(t) = 20e0.02t?

(d) If r(t) is as in part ( c), and if the tank con-tains 3000 gallons initially, find a formulafor Q(t), the amount of water in the tankat time t.

(a) Amount of water entering tank in a short period oftime = Rate×Time = r(t)∆t.

(b) Summing the contribution from each of the smallintervals ∆t:

Amount of water entering the tank

≈∑i

r(ti)∆ti

where ∆t = 5/n. Taking a limit as ∆t→ 0, we getthe integral form instead of the sum:

Exact amount of water entering the tank

=

∫ 5

0

r(t) dt.

(c) If Q(t) is the amount of water in the tank at time t,then Q′(t) = r(t). We want to calculate net changein volume between t = 0 and t = 5, or Q(5)−Q(0).By the Fundamental Theorem,

Amount which has entered tank

= Q(5)−Q(0)

=

∫ 5

0

r(t) dt

=

∫ 5

0

20e0.02t dt

=20

0.02e0.02t

∣∣∣50

= 1000(e(0.02)(5) − 1) ≈ 105.17 gallons.

(d) By the Fundamental Theorem again,

Amount which has entered tank

= Q(t)−Q(0)

=

∫ t

0

r(t) dt

Q(t)− 3000 =

∫ t

0

20e0.02t dt

so Q(t) = 3000 +

∫ t

0

20e0.02u du

Note: t is already being used, so we put u insidethe integral; since this is a definite integral, thevariable inside the integral will disappear when wesub in the limits.

Q(t) = 3000 +20

0.02e0.02t

∣∣∣t0

= 3000 + 1000(e0.02t − e0)

= 1000e0.02t + 2000.

This is the quantity of water in the tank at timet. Note that we can do a basic sanity check onthe answer by verifying that Q(0) = 3000 (given),which is true for the formula we arrived at:

Q(0) = 1000e0.02·0 + 2000 = 3000

15

73. After a spill of radioactive iodine, measurementsat t = 0 showed the ambient radiation levels atthe site of the spill to be four times the max-imum acceptable limit. The level of radiationfrom an iodine source decreases according to theformula

R(t) = R0e−0.004t

where R is the radiation level (in millirems/hour) at time t in hours and R0 is the initialradiation level (at t =0).

(a) How long will it take for the site to reachan acceptable level of radiation?

(b) Engineers look up the safe limit of radiationand find it to be 0.6 millirems/hour. Howmuch total radiation (in millirems) will havebeen emitted by the time found in part (a)?

(a) If the level first becomes acceptable at time t1, thenR0 = 4R(t1), and

1

4R0 = R0e

−0.004t1

1

4= e−.004t1

Taking natural logs on both sides yields

ln

(1

4

)= −0.004t1

t1 =ln(

14

)−0.004

t1 ≈ 346.574 hours.

(b) Since the initial radiation was four times the ac-ceptable limit of 0.6 millirems/hour, we have R0 =4(0.6) = 2.4. The rate at which radiation is emit-ted is R(t) = R0e

−.004t, so

Total radiation emitted =

∫ 346.574

0

2.4e−0.004t dt.

Finding by guess-and-check or substitution that an

antiderivative of e−0.004t ise−0.004t

−0.004,

∫ 346.574

0

2.4e−0.004t dt

= 2.4e−0.004t

−0.004

∣∣∣346.574

0

= 2.4

[e−0.004(346.574)

−0.004− e0

−0.004

]= 450.00

We find that 450 millirems were emitted duringthis time.

74. David is learning about catalysts in his Chem-istry course. He has read the definition:

Catalyst: A substance that helps a re-action to go faster without being usedup in the reaction.

In today’s Chemistry lab exercise, he has to adda catalyst to a chemical mixture that producescarbon dioxide. When there is no catalyst, thecarbon dioxide is produced at a rate of 8.37 ×10−9 moles per second. When C moles of thecatalyst are present, the carbon dioxide is pro-duced at a rate of (6.15× 10−8)C + 8.37× 10−9

moles per second.

The reaction begins at exactly 10:00 a.m. Oneminute later, at 10:01 sharp, David starts to addthe catalyst at a constant rate of 0.5 moles persecond.

How much carbon dioxide is produced between10:00 (sharp) and 10:05?

In the first minute of the reaction no catalyst is presentand the amount of carbon dioxide produced is

(8.37× 10−9mol/s)(60s) = 5.02× 10−7mol.

In the next four minutes, catalyst is being added. Af-ter t seconds of adding catalyst of 0.5mol/s, there is0.5tmol of catalyst present. Thus, at time t carbondioxide is being produced at a rate of

(6.15× 10−8)(0.5t) + (8.37× 10−9mol/s)

= (3.075× 10−8)t+ (8.37× 10−9mol/s).

During the four minutes when catalyst is being added,the amount of carbon dioxide produced is:

∫ 240

0

(3.076× 10−8)t+ (8.37× 10−9)dt

= (1.5375× 10−8)t2 + (8.37× 10−9)t∣∣∣240

0

= 8.88× 10−4mol

The total amount of CO2 produced in the first fiveminutes of the reaction is therefore

5.02× 10−7 + 8.88× 10−4 = 8.88× 10−4mol.

16

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