Quiz 6 Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a ? are ‘‘for your entertainment’’ and are not essential. SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference sections 6.6 and 7.2 of the recommended textbook (or the equivalent section in your alternative textbook/online resource) and your lecture notes. EXPECTED SKILLS: • Be able to use integration to compute the work done by a variable force in moving an object along a straight path from x = a to x = b. Specifically, be able to solve spring problems, lifting problems, and pumping problems. • Be able to use integration by parts to evaluate various integrals, in- cluding integrands involving products of functions, isolated logarithmic functions, or isolated inverse trigonometric functions. 1
Transcript
Quiz 6 Practice Problems
Practice problems are similar, both in difficulty and in scope, to the type of
problems you will see on the quiz. Problems marked with a ? are ‘‘for your
entertainment’’ and are not essential.
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference sections
6.6 and 7.2 of the recommended textbook (or the equivalent section in your
alternative textbook/online resource) and your lecture notes.
EXPECTED SKILLS:
• Be able to use integration to compute the work done by a variable
force in moving an object along a straight path from x = a to x = b.
Specifically, be able to solve spring problems, lifting problems, and
pumping problems.
• Be able to use integration by parts to evaluate various integrals, in-
cluding integrands involving products of functions, isolated logarithmic
functions, or isolated inverse trigonometric functions.
1
PRACTICE PROBLEMS:
1. A variable force F (x) is applied in the positive x direction, as shown in
the graph below.
Find the work done by the force in moving a particle from x = 0 to
x = 9.
Spring Problems
2. A force of 20 lb is required to hold a spring stretched 5 in beyond its
natural length. How much work is done in stretching the spring from
its natural length to 7 in beyond its natural length.
3. A spring has a natural length of 2 ft. A body of 10 lb hanging on the
spring stretches it to a total length of 2.4 ft.
(a) Find the spring constant (in pounds per foot).
(b) How far beyond its natural length would a body of 50 lb stretch
the spring?
(c) How much work is required to stretch the spring from its natural
length to a length of 3 ft?
2
4. If the work required to stretch a spring 1 foot beyond its natural length
is 15 ft-lb, how much work is needed to stretch it 8 inches beyond its
natural length?
Lifting Problems
5. A cable that weighs 3 lb/ft is used to lift 1,000 lb of iron ore up a
mineshaft which is 450 ft deep. Find the work done.
6. A 5 lb bucket containing 10 lb of water is hanging at the end of a 30 ft
rope which weighs 1/2 lb/ft. The other end of the rope is attached to a
pulley. Assume that the rope is wound onto the pulley at a rate of 3
ft/s, causing the bucket to be lifted. Find the work done in winding
the rope onto the pulley.
7. Repeat problem 6 adding the assumption that water leaks out of the
bucket at a rate of1
4lb/s.
8. A bag of sand originally weighing 144 lb is lifted at a constant rate. As
it rises, sand also leaks out at a constant rate. At the instant when the
bag has been lifted to a height of 18 feet, exactly half of the original
amount of sand remains. (Neglect the weight of the bag and the lifting
equipment.)
(a) Describe the weight of the sand as a function of x, where x is the
height of the bag above the ground.
(b) Calculate the work done in lifting the sand to the height of 18 ft
from the ground.
3
Pumping Problems
9. A circular swimming pool has a diameter of 10 meters. The sides are
4 meters high. And, the depth of the water is 3.5 meters. How much
work is required to pump all of the water over the side? Recall that
water weighs 9810 N/m3.
10. At Charlie’s Chocolate Factory, a tank in the shape of an inverted
circular cone having a height of 10 meters and a radius (at the top) of 6
meters is filled with chocolate pudding to a height of 2 meters. In order
to sterilize the tank, the factory needs to empty the tank. Find the
work required to empty the tank by pumping the chocolate pudding
through a hole in the top of the tank. Note: the weight density of
chocolate pudding is 12,178 N/m3.
11. The trough pictured below is 15 feet long and 4 feet wide at the top.
The ends of the trough are isosceles triangles with a height of 3 feet.
If the trough is filled to a height of 1 foot with water, find the work
required to pump all of the water over the side. Recall that water
weighs 62.4 lb/ft3.
4
12. The tank shown below is half full of oil which has a weight-density of
9016 N/m3.
Let x = 0 correspond to the bottom of the tank. Set up an integral
which represents the work (in joules) required to pump the oil out of
the outlet at the top which is one meter above the top of the tank. Do
not evaluate your integral.
For problems 13-22, evaluate the given integral.
13.
∫xe4x dx
14.
∫x2 cosx dx
15.
∫x2 lnx dx
16.
∫lnx
x4dx
17.
∫sin–1x dx
18.
∫x sec2x dx
19.
∫e2x cos 3x dx
5
20.
∫x3 cos(x2) dx
21.
∫ π
0
3x sinx dx
22.
∫ 1
0
x2ex dx
23. Suppose that u and v are differentiable functions of x with
∫ 1
0
v du = 3
and the following functional values.
x u(x) v(x)
0 5 2
1 7 –4
Use this information to compute
∫ 1
0
u dv.
24. Evaluate
∫sin√x dx by first making an appropriate substitution and
then applying integration by parts.
25. Evaluate
∫(sin–1x)2 dx.
26. Find the area of the region which is enclosed by y = lnx, y = 1, and
x = e2.
6
27. Let R be the region enclosed by the graphs y = lnx, x = e, and the
x-axis (as shown below).
Find the volume of the solid that results from revolving R around the
line y = –1.
28. Let f be a differentiable function. Use integration by parts to show:
∫f(x) dx = xf(x)−
∫xf ′(x) dx.
? The Great Pyramid of King Khufu was built of limestone in Egypt over
a 20-year time period from 2580 bc to 2560 bc. Its base is a square with
side length 756 ft and its height when built was 481 ft. (It was the tallest
man-made structure in the world for more than 3800 years.) The density of
the limestone is about 150 lb/ft3.
(a) Estimate the total work done in building the pyramid.
(b) If each laborer worked 10 hours a day for 20 years, for 340 days a year,
and did 200 ft-lb/h of work in lifting the limestone blocks into place, about
7
how many laborers were needed to construct the pyramid?
? Find an antiderivative of f(x) = |x| that is not defined piecewise.
? Compute the integral ∫(1 + 2x2)ex
2
dx.
8
SOLUTIONS
1. Here W =
∫ 9
0
F (x) dx and corresponds to the area under the trapezoid,
which is1
2h(b1 + b2) =
1
2(6 lb)(9 ft + 5 ft) = 42 ft-lb.
2. Recall from Hooke’s Law that F (x) = kx where k is the yet-to-be-
determined spring constant. We are given that F (5) = 20, and so
5k = 20 ⇒ k = 4 lb/in. The desired work done is thus
∫ 7
0
F (x) dx =∫ 7
0
4x dx = 2x2∣∣∣∣70
= 2(7)2 in-lb = 98 in-lb =49
6ft-lb.
3. (a) The 10 lb body stretches the spring 0.4 ft beyond its natural length,
and so F (0.4) = 10. Hooke’s Law dictates that 0.4k = 10⇒ k =
10/0.4 lb/ft = 25 lb/ft.
(b) Using Hooke’s Law, we need 50 = 25x⇒ x = 2 ft.
(c) If the spring stretches to 3 ft, then it is 1 ft beyond its natural
length. Thus W =
∫ 1
0
F (x) dx =
∫ 1
0
25x dx =25
2x2∣∣∣∣10
=25
2ft-lb.
4. Based on what is given and Hooke’s Law, we see that, if k is the spring
constant, then 15 =
∫ 1
0
F (x) dx =
∫ 1
0
kx dx =k
2x2∣∣∣∣10
=k
2⇒ k =
30 lb/ft. Since 8 inches is 2/3 feet, the desired work is
∫ 3/4
0
F (x) dx =∫ 3/4
0
30x dx = 15x2
∣∣∣∣∣2/3
0
= 15
(2
3
)2
ft-lb =20
3ft-lb.
5. Let x be the distance from the bottom of the shaft. Notice F (x) =
Fore(x) + Fcable(x). Certainly Fore(x) = 1000. Now, when the ore is x
feet from the bottom of the mineshaft, there is 450− x feet of rope left
to pull, which weighs 3(450− x) lb. Hence F (x) = 1000 + 3(450− x) =
9
2350−3x, and so the work done is
∫ 450
0
F (x) dx =
∫ 450
0
(2350−3x) dx =[2350x− 3
2x2]4500
= (2350)(450)− 3
2(450)2 = 753, 750 ft-lb.
6. Let x be the distance that the water has been lifted. Notice F (x) =
Fbucket(x)+Fwater(x)+Frope(x). Certainly Fbucket(x) = 5 and Fwater(x) =
10. Now, when the bucket has been lifted x feet, there is 30−x feet of rope
left to pull, which weighs1
2(30−x) lb. Hence F (x) = 5+10+
1
2(30−x) =
30− 1
2x, and so the work done is
∫ 30
0
F (x) dx =
∫ 30
0
(30− 1
2x
)dx =[
30x− 1
4x2]300
= (30)(30)− 1
4(30)2 = 675 ft-lb.
7. The only thing that changes here is Fwater(x). Since the rope is wound
onto the pulley at a rate of 3 ft/s, when the bucket has been lifted x
feet, x/3 seconds have elapsed, and sox
3· 14=
1
12x pounds of water
have leaked from the bucket. Thus Fwater(x) = 10 − 1
12x, and so
F (x) = 5 + 10 − 1
12x + 15 − 1
2x = 30 − 7
12x. The work done is now∫ 30
0
F (x) dx =
∫ 30
0
(30− 7
12x
)dx =
[30x− 7
24x2]300
= (30)(30) −
7
24(30)2 =
1275
2ft-lb.
8. (a) Let F (x) be the weight of the sand as a function of x, and let
c be the rate at which sand leaks from the bag (in lb/ft, which
is allowable since the bag is lifted at a constant rate). Then
F (x) = 144 − cx. We are given that F (18) = 72 so that 72 =
144− 18c⇒ 18c = 72⇒ c = 4. Therefore F (x) = 144− 4x.
(b) W =
∫ 18
0
F (x) dx =
∫ 18
0
(144−4x) dx = [144x− 2x2]180 = (144)(18)−
2(18)2 = 1944 ft-lb.
10
9. Consider a thin circular disk of water of thickness dx that is x meters
from the bottom of the pool. Note that such a disk has to move a
distance of 4−x meters. Let dW be the work done to move this disk of
water. Then dW = h(x) dF where h(x) = 4− x and dF is the weight
of the disk. Then dF = ρ dV where ρ is the density of water and dV is
the volume of the disk. As usual, dV = A(x) dx where A(x) is the cross-
sectional area. Observe that the cross-sections are circles, and since
the pool is presumably cylindrical, we have that A(x) = π(5)2 = 25π
since the radius of the pool is 5 meters (half of the diameter). Hence
W =
∫ 3.5
0
h(x) dF =
∫ 3.5
0
(4 − x)ρ dV =
∫ 3.5
0
(4 − x)(9810)A(x) dx =∫ 3.5
0
(9810)(25π)(4− x) dx ≈ 6.0675× 106 J.
10. Consider a thin circular disk of pudding of thickness dx that is x meters
from the bottom of the tank. Note that such a disk has to move a
distance of 10− x meters. Let dW be the work done to move this disk
of pudding. Then dW = h(x) dF where h(x) = 10 − x and dF is the
weight of the disk. Then dF = ρ dV where ρ is the density of chocolate
pudding and dV is the volume of the disk. As usual, dV = A(x) dx,
where A(x) is the cross-sectional area. Observe that the cross-sections
are circles so that A(x) = π[r(x)]2 where r(x) is the radius of the
circular cross-section at a height of x meters. We can compute r(x)
from similar triangles as follows:
r(x)
x=
6
10=
3
5⇒ r(x) =
3
5x.
(Draw a picture to see why.) Therefore A(x) = π
(3
5x
)2
=9
25πx2.
HenceW =
∫ 2
0
h(x) dF =
∫ 2
0
(10−x)ρ dV =
∫ 2
0
(10−x)(12,178)A(x) dx =
11
∫ 2
0
(12,178)
(9
25πx2)(10− x) dx ≈ 2,484,312
25π J.
11. Consider a thin rectangular lamina of water of thickness dx that is
x feet from the bottom of the tank. Note that such a lamina has to
move a distance of 3− x feet. Let dW be the work done to move this
lamina of water. Then dW = h(x) dF where h(x) = 3 − x and dF is
the weight of the lamina. Then dF = ρ dV where ρ is the density of
water and dV is the volume of the lamina. As usual, dV = A(x) dx,
where A(x) is the cross-sectional area. Observe that the cross-sections
are rectangles so that A(x) = l(x)w(x), where l(x) and w(x) are the
length and width, respectively, of the rectangular cross-section at a
height of x feet. Observe that l(x) is a constant 15 ft. We can compute
w(x) from similar triangles as follows:
w(x)
x=
4
3⇒ w(x) =
4
3x.
(Draw a picture to see why.) Therefore A(x) = 15
(4
3x
)= 20x.
Hence W =
∫ 1
0
h(x) dF =
∫ 1
0
(3− x)ρ dV =
∫ 1
0
(3− x)(62.4)A(x) dx =∫ 1
0
(62.4)(20x)(3− x) dx = 1456 ft-lb.
12. Consider a thin rectangular lamina of oil of thickness dx that is x feet
from the bottom of the tank. Note that such a lamina has to move a
distance of 1+6−x = 7−x meters. Let dW be the work done to move
this lamina of oil. Then dW = h(x) dF where h(x) = 7− x and dF is
the weight of the lamina. Then dF = ρ dV where ρ is the density of oil
and dV is the volume of the lamina. As usual, dV = A(x) dx, where
A(x) is the cross-sectional area. Observe that the cross-sections are
rectangles so that A(x) = l(x)w(x), where l(x) and w(x) are the length
12
and width, respectively, of the rectangular cross-section at a height of
x feet. Observe that l(x) is a constant 8 m. We can compute w(x)
using circular geometry. (I know that this will be difficult to follow, so
definitely draw a picture to guide you through the following argument.)
Indeed, the long sides of the lamina are chords of the circular sides of
the tank. Let r(x) be the distance from the center of a circular side
of the tank to either end of such a chord on the same side of the tank.
(This length is the same for both ends since the segment connecting
them is a radius of the circle.) Since r(x) is merely the radius of the
circle, we see that r(x) is a constant 3 m. The perpendicular distance
from the center of the circle to the chord is 3− x meters and, by the
property of isosceles triangles, bisects the chord. By the Pythagorean
Theorem,
32 = (3−x)2+[w(x)
2
]2=
[w(x)]2
4= 9−(3−x)2 ⇒ w(x) =
√4(9− (3− x)2).
Therefore A(x) = 8 ·2√
9− (3− x)2. HenceW =
∫ 3
0
h(x) dF =
∫ 3
0
(7−
x)ρ dV =
∫ 3
0
(7−x)(9016)A(x) dx =
∫ 3
0
(9016)(7−x)(16√9− (3− x)2
)dx =
144,256
∫ 3
0
(7− x)√6x− x2 dx J.
13
13. Using tabular integration
u dv SIGN
x e4x +
11
4e4x −
01
16e4x +
the answer is1
4xe4x − 1
16e4x + C.
14. Using tabular integration
u dv SIGN
x2 cosx +
2x sinx −
2 – cosx +
0 – sinx −
the answer is x2 sinx+ 2x cosx− 2 sinx+ C.
15. Note that the integral is the same as
∫(lnx)(x2) dx. Since lnx is
logarithmic and x2 is algebraic, LIATE suggests that we let u = lnx
and dv = x2 dx. Then du = (1/x)dx and v = (x3/3)dx. Hence
14
∫x2 lnx dx =
∫u dv = uv−
∫v du = (lnx)
(x3
3
)−∫x3
3·1xdx =
1
3x3 lnx−1
9x3+C.
16. Note that the integral is the same as
∫(lnx)(x–4) dx. Since lnx is
logarithmic and x–4 is algebraic, LIATE suggests that we let u = lnx
and dv = (x–4) dx. Then du = (1/x)dx and v = –(x–3/3) = –1/(3x3).
Hence
∫lnx
x4dx =
∫u dv = uv−
∫v du = (lnx)
(–
1
3x3
)−∫ (
–1
3x3
)(1
x
)dx.
Note that the remaining integral above is positive. Now
∫ (1
3x3
)(1
x
)dx =
∫1
3x4dx = –
1
9x3+ C,
and so the answer is
–lnx
3x3− 1
9x3+ C.
17. Use the standard trick for integrating isolated inverse trigonometric
functions (that is, multiply by 1). Let u = sin–1x and dv = dx. Then
du =1√
1− x2dx and v = x. So
∫sin–1 x dx =
∫u dv = uv−
∫v du = (sin–1x)(x)−
∫(x)
(1√
1− x2
)dx.
15
To find
∫x√
1− x2dx, let w = 1− x2 so that dw = –2x dx⇒ –
1
2dw =
x dx. Hence
∫x√
1− x2dx = –
1
2
∫dw√w
= –1
2· 2√w + C =
√1− x2 + C.
The final answer is thus
x sin–1x+√1− x2 + C.
18. Using tabular integration
u dv SIGN
x sec2x +
1 tanx −
0 ln|secx| +
the answer is x tanx− ln|secx|+ C = x tanx+ ln|cosx|+ C.
16
19. We use tabular integration.
u dv SIGN
e2x cos 3x +
2e2x1
3sin 3x −
4e2x –1
9cos 3x +
∫
We stop at the last row since it is a constant multiple of the first. The
table tells us that
∫e2x cos 3x dx =
1
3e2x sin 3x+
2
9e2x cos 3x− 4
9
∫e2x cos 3x dx.
Adding4
9
∫e2x cos 3x dx to both sides yields
13
9
∫e2x cos 3x dx =
1
3e2x sin 3x+
2
9e2x cos 3x.
Multiply both sides by 9/13 so that
∫e2x cos 3x dx =
3
13e2x sin 3x+
2
13e2x cos 3x+ C.
20. First let w = x2 so that dw = 2x dx⇒ 12dw = x dx. Hence
∫x3 cos(x2) dx =
∫x2 cos(x2) · x dx =
1
2
∫w cosw dw.
17
This new integral can be solved with integration by parts. Using tabular
integration
u dv SIGN
w cosw +
1 sinw −
0 – cosw +
this new integral is w sinw + cosw. Plugging back in w = x2 and
