Chapter 6. Applications of Integration

Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

http://matrix.skku.ac.kr/Cal-Book1/Ch5/

Chapter 6. Applications of Integration

6.1 Areas between Curves   http://youtu.be/o53phm5cqJE

6.2 Volumes   http://youtu.be/4-ChOAFbJAs

문제풀이 by 김종민  http://youtu.be/Fd4Mguf2dbU

6.3 Volumes by Cylindrical Shells   http://youtu.be/qM1izf8qeX8

문제풀이 by 신영찬  http://youtu.be/gNaKkA0UNHg

6.4 Work   http://youtu.be/u3ZaJWhKy6k

문제풀이 by 김건호  http://youtu.be/SmIo2yaxNsY

6.5 Average Value of a Function   http://youtu.be/zmEeGmwQTB0

문제풀이 by 신종희  http://youtu.be/BVahd-DJoe8

6.1 Area between Curves

As mentioned in Chapter 5, if is a function on an integral such that for all in , then is the area under the graph of .

Consider two continuous functions and defined on an interval such that . Let be the region that lies between the two given curves and and between the vertical lines and . (See Figure 1.) We want to derive a formula for the area of .

Suppose for a moment that for all in . Then, if we denote by the region under the graph of and by the region under the graph of , then we can easily see that

.

In the general case, when and are not necessarily non-negative, we can see that this formula remains true by shifting vertically both the graphs of and of by the same distance, so that both will be above the -axis.

Figure 1

The area of the region bounded by the curves and on , and the lines and is

[1]    [area under ][area under ]

Figure 2

In the special case when , is the region under the graph of (and above the -axis). In general, if and are continuous functions with on the interval (see Figure 3), then the area of the region bounded by the curves , on , the lines and is

.

Figure3

EXAMPLE 1

Determine the area of the region bounded above by , bounded below by , and bounded on the sides by and .

Solution.  Draw a typical approximating rectangle with width (see Figure 4), with the upper boundary curve and the lower boundary curve , , and .

Figure 4

f = exp(x)

p = plot(f, 0, 1, fill=x/2) + plot(f, 0, 1.5)+plot(x/2, 0, 1.5)

l1=line([(1, 1/2),(1, exp(1))], thickness=1.2, linestyle="--", color='red')

l2=line([(0, 0), (0, exp(1))], thickness=2, linestyle="--", color='red')

t1=text("$y=e^x$", (0.7, 3), fontsize=50, color='blue')

t2=text("$y= x/2$", (1.2, 0.3), fontsize=50, color='blue')

show(p+l1+l2+t1+t2)

integral(e^x - (1/2)*x, x, 0, 1)

EXAMPLE 2

Calculate the area of the region enclosed by and .

Solution. First we need to find the intersection points of the line and parabola , in order to know the bounds of the integral. The coordinate of the intersection points satisfies both and , hence it satisfies , or . Thus , so or . The points of intersection are and . From Figure 5, the top and bottom boundaries are and respectively.

The area of a typical rectangle is and the region lies between and . So the total area is

Figure 5

If for some values of but for other values of , then the area between the curves and is obtained by splitting the given region into several regions. In Figure 6,  the region is obtained by adding areas , , of  , , . We will use the following “word” formulas to ensure that we remember that the area is always the “larger” function minus the “smaller” function.

Figure 6

Then the area of the region is defined to be the sum of the areas of the smaller regions  , , that is, . The area of the region is defined to be the sum of the areas of the smaller regions , , , that is, ,

Since

[2]                  ,

the area between the curves and and between and is given by

.

EXAMPLE 3

Compute the area of the region bounded by the curves , , and .

http://matrix.skku.ac.kr/cal-lab/cal-6-1-Exm-3.html

Solution.  The points of intersection of the curves and occur when , that is, when (since ). (See Figure 7.)

Figure 7

Note that when but when . Therefore, the required area is

f = x^2

g = (x-2)^2

integral(abs( f- g), x, 0, 2)

(The region is symmetric about , so .)

Suppose that and are continuous functions with on the interval . The area of the region bounded by the curves , on , and the lines , is

. (See Figure 8 and 9.)

Figure 8                     Figure 9

EXAMPLE 4

Find the area enclosed by the parabola and the line .

http://matrix.skku.ac.kr/cal-lab/cal-6-1-Exm-4.html

Solution. The points of intersection and are obtained by solving the given two equation simultaneously. The region of interest lies between and and with left and right boundary curves (see Figure 10) given by, respectively:

.

Thus

Figure 10

var('x, y')

f(x) = 2*y^2

g(x) = y+1

solve([f==g], y)

integral(abs( f- g), y, -1/2, 1).n()

Answer :  {9} over {8} =1.12500

Caution In this example, if we try to find the area by integrating with respect to instead of with respect to , the region needs to be split into two areas labeled and (see Figure 10) with more calculations involved.

CAS EXAMPLE 5

Determine the area of the region bounded by the curves:

.

Solution.

Figure 11

integral(x - 1/x, x, 1, 2)

CAS EXAMPLE 6

Find the area between the curves and between and .

http://matrix.skku.ac.kr/cal-lab/cal-6-1-example-6.html

Solution.

Figure 12

var('x')

f(x)= 2*x^2 +10

g(x)= 4*x +16

L1=line([(-2,0),(-2,50)], color="red", linestyle='--')

L2=line([(-5,-10),(-5,80)], color="red", linestyle='--')

P1=plot(f,(x,-6,-1))

P2=plot(g,(x,-6,-1), color="green")

show(P1+P2+L1+L2)

integral((2*x^2 +10) - (4*x +16), x, -5, -2)

6.1 EXERCISES (Areas between Curves)

1-21. Find the area of the region, bounded by the given curves.

1.

Solution.

f(x)=abs((5-x^2)-(x-1))

plot(f(x),x,0,2)

print integral(f(x),x,0,2)

2.

Solution.

f(x)=exp(x)-cos(x)

plot(f(x),x,0,pi/2)

print integral(f(x),x,0,pi/2)

3.

Solution.

f(x)=sqrt(x)-x

plot(f(x),x,0,1)

4.

Solution.

5.

Solution.

6.

Solution.

7.

Solution.

8.

Solution.

9.

Solution.

10.

Solution.

11.

Solution.

12.

Solution.

has solutions ,

so is positive where

var('x')

f(x)=sqrt(x)

g(x)=-sqrt(x)

h(x)=-x+2

P = plot(f,x, xmin=0, xmax=5, ymax=3, ymin=-3, color='blue')

Q = plot(g,x, xmin=0, xmax=5, ymax=3, ymin=-3, color='green')

R = plot(h,x, xmin=-1, xmax=5, ymax=3, ymin=-3, color='red', linestyle="--")

show(P+Q+R)

F(x)=(x+2)-x^2

print integral(F(x), x, -1,2)    # 9/2

plot(F(x), x, -1, 2)

13.

Solution.

14.

Solution.

15. ,

Solution.

16.

Solution.

17.

Solution.

18.

Solution.

19.

Solution.

20.

Solution.

21.

Solution.

22-23. Find the integral and interpret it as the area of a region.

22.

Solution.

23.

Solution.

24-25. Approximate the area of the region bounded by the given curves using the Midpoint Rule with .

24.

Solution.

25.

Solution.

var('x')

f(x) = x - sqrt (9 - x^(2) )

integral(f, x, 0, 3)

26-29. Determine the area of the region bounded by the curves.

26.

Solution.

27.

Solution.

28.

Solution.

29.

Solution.

30. Find the area of region defined by the inequalities .

Solution.

31. Find the area enclosed by the loop of the curve with equation .

(Tschirnhausen's cubic.)

Solution.

32. Find the area of the region bounded by the curve , the tangent line to this curve at , and the -axis.

Solution.

var('x')

f(x)=12*x-16

g(x)=x^3

print integral(g(x), x, 0, 4/3)+ integral(g(x)-f(x), x, 4/3,2)

plot(f(x), 0,3, ymax=15)+plot(g(x), 0, 3, ymax=15, color='red')

33. Find the number such that the line divides the region bounded by the curves and into two regions with equal area.

Solution.

34. (a) Find the number such that the line bisects the area under the curve

.

(b) Find the number such that the line bisects the area in part (a).

Solution.

35. Find the values of such that the area of the region enclosed by the parabolas and is 1944.

Solution.

For and is another solution.

Therefore, .

36-40. Find the area of the region bounded by the given curves.

36. , .

Solution. Since ,

.

37. , .

Solution. Since

38. , .

Solution. Since

39. ,

Solution.

Solution.

var('x')

f(x) = sqrt (2*x)

g(x) = x^2

integral(g-f, x, 2^(1/3), 2)

6.2 Volume

Consider a solid called a cylinder (or, more precisely, a circular cylinder) [see Figure 1(b)]. A cylinder is bounded by a plane region , called the base, and a congruent region in a parallel plane [see Figure 1(a)]. The cylinder consists of all points on line segments that are perpendicular to the base and join to . If the area of the base is and the height of the cylinder (the distance from to ) is , then the volume of the cylinder is . In particular, if the base is a circle with radius , then the cylinder is a circular cylinder with volume [see Figure 1(b)], and if the base is a rectangle with length and width , then the cylinder is a rectangular box (also called a rectangular parallelepiped) with volume [see Figure 1(c)].

Figure 1

A cross-section of a solid is a plane region obtained by intersecting with a plane. Let be the area of the cross-section of in a plane perpendicular to the -axis and passing through the point , where .(See Figure 2.) The cross-sectional area will vary as increases from to .

Figure 2

Divide into “slabs” of equal width using the planes , , to slice the solid. Choose sample points in . Approximate the volume of slab (the part of that lies between the planes and ) by a cylinder with base area and “height” . (See Figure 3.)

Figure 3

The volume of this cylinder is . So an approximation of the volume of the slab is . Adding the volumes of these slabs, we get an approximation to the total volume . As the volume is the limit of a Riemann sum.

DEFINITION of Volume

Let be a solid that lies between and . If the moving cross-sectional area of in the plane , through and perpendicular to the -axis, is , where is a continuous function, then the volume of is

.

For a cylinder, the cross-sectional area is constant: for all R. So, the definition of volume for a cylinder gives , which agrees with the formula .

Solid of Revolution

A solid of revolution is a solid obtained by revolving a plane region about a line. (See Figure 4.) The volume of a solid of revolution is calculated using the formula if the cross-sections are perpendicular to the -axis

or if the cross-sections are perpendicular to the -axis

.

Note that the cross-section is either a disk or a ring, as shown in Figure 4. If it is a disk with radius , then . If it is a ring with outer radius and inner radius , then . Thus, if is the region bounded by the curve , the lines , , and the -axis, then    ,

hence the volume of the solid obtained by revolving the region about the -axis is

. (See Figure 4 (a), (b).)

If is region bounded by , , and the lines , , with then the volume of the solid obtained by revolving the region about the -axis is

. (See Figure 4 (c).)

(a)

(b)

(c)

EXAMPLE 1

Calculate the volume of the solid obtained by rotating the region bounded by the curves and about the line and about the -axis. [See Figure 4(b).]

Solution. The cross-sectional area is and the volume of the solid is .

EXAMPLE 2

Find the volume of a sphere of radius .

Solution. Assume that the center of the sphere is at the origin. (See Figure 5.) Then the plane intersects the sphere in a circle whose radius is So the cross-sectional area is

.

From the definition of volume with and , we have

.

Figure 5

EXAMPLE 3

Compute the volume of the solid obtained by rotating the region bounded by , and about the -axis.

http://matrix.skku.ac.kr/cal-lab/cal-6-2-Exm-9.html

Solution. The region bounded by , and is shown in Figure 6(a). The resulting solid of revolution obtained by rotating this region about the -axis is shown in Figure 6(b). Since the region is rotated about the -axis, slice the solid perpendicular to the -axis and integrate with respect to . A  slice at height is a circular disk with radius , where . So the area of a cross-section through is and the volume of the approximating cylinder [shown in Figure 6(b)] is . Since the solid lies between and , its volume is

.

(a)                         (b)

Figure 6

r = var('r')

f(r) = pi* r^2

integral(f, r, 0, r)

EXAMPLE 4

Calculate the volume of the solid obtained by rotating about the -axis the region under the curve from to .

Solution. The region under the curve from to is shown in Figure 7(a). By rotating this region about the -axis, the solid shown in Figure 7(b) is obtained. A slice through the point , gives a disk with radius . Its area of cross-section is and its thickness is .

The solid lies between and , so its volume is

.

Figure 7

EXAMPLE 5

Find the volume of the solid of revolution obtained by revolving the region enclosed by the curves and about the line .

Figure 8

Solution. The cross-section is a washer with the inner radius and the outer radius . (See Figure 8.) The cross-sectional area is

and thus the volume of is given by

EXAMPLE 6

Find the volume of the solid of revolution obtained by rotating the region in Example 4 about the line .

Solution. A horizontal cross-section of a washer with inner radius and outer radius is shown in Figure 9. Its cross-sectional area is

The volume is

Figure 9

Examples of solids that are not solids of revolution:

We can also use the definite integral to find the volume of a solid with specific cross sections on an interval, if we know a formula for the region determined by each cross section. If the cross sections generated are perpendicular to the -axis, then their areas will be functions of , denoted by . The volume () of the solid on the interval is

EXAMPLE 7

Find the volume of the solid of revolution obtained by rotating the region in Example 4 about the line .

Solution. Place the origin at the vertex of the pyramid and the -axis along its central axis as in Figure 10. Any plane that passes through and is perpendicular to the -axis intersects the pyramid in a square with side of length , say. Express in terms of by observing from the similar triangles in Figure 11 that

and so . [or observe that the line has slope and so its equation is .] Thus, the cross-sectional area is

.

The pyramid lies between and , so its volume is

.

Figure 10                                                  Figure 11

EXAMPLE 8

A wedge is cut out of a circular cylinder of radius 3 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of along a diameter of the cylinder. Find the volume of the wedge.

Solution. Align the -axis along the diameter where the planes meet. Then the base of the solid is a semicircle with equation , . A cross-section perpendicular to the -axis at a distance from the origin is a triangle , as shown in Figure 12, whose base is and whose height is . (See Figure 12.) Thus, the cross- sectional area is

and the volume is

Figure 12

CAS EXAMPLE 9

Calculate the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

Solution.

var('x, y')

p1 = parametric_plot((2, y), (y, -1, 3), color='blue')

p2 = parametric_plot((4, y), (y,-1, 3), color='green')

p3 = implicit_plot(x*y-4==0, (x, 0.5, 5), (y, -1, 3))

p4 = parametric_plot((x, 0), (x, 0.5, 5), color='red')

show(p1+p2+p3+p4)

integral(pi*(4/x)^2, x, 2, 4)

Figure 13

6.2 EXERCISES (Volumes)

1-17. Calculate the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

Solution.

var('x, y')

f(x)=pi*(1/x)^2

print integral(f(x), x, 1, 3)            # 2/3 *pi

p0 = plot(f(x), x, 0, 2, ymax=30)

p1 = parametric_plot((1, y), (y, -1, 3), color='blue')

p2 = parametric_plot((3, y), (y,-1, 3), color='green')

show(p0+p1+p2)

Solution.

var('x, y')

f(x)=pi*(x^3)^2

print integral(f(x), x, 0, 2)

g=x-y^2

p1=plot(x^3, x, -1, 3)

p2=parametric_plot((x,0), (x, -1, 3), color='green')

p3=parametric_plot((2,y), (y, -1, 25), color='red')

show(p1+p2+p3)

var('x, y, z')

implicit_plot3d(x^2 + z^2 == y^(6) , (x,-2, 2), (y,0, 8), (z,-2, 2), opacity=0.5, color="red")

Solution.

var('x, y')

f(x)=pi*(1/x^2)^2

print integral(f(x), x, 1, 2)

plot(f(x), x, 0, 2, ymax=50)

Solution.

var('x, y')

p1 = parametric_plot((2,y), (y, -1, 5), color='blue')

p2 = parametric_plot((5,y), (y,-1,5), color='green')

p3 = implicit_plot(y==sqrt(x^2-1), (x, 1, 6), (y, -1, 5))

p4 = parametric_plot((x,0), (x, 0.5, 6), color='red')

show(p1+p2+p3+p4)

print integral

Solution.

var('x, y')

f(x)=pi*(81-x^4)

print integral(f(x), x, 0, 3)

p1=plot(x^2, x, 0, 3)

p2=parametric_plot((x,9), (x, -1, 4), color='green')

p3=parametric_plot((0,y), (y, -1, 10), color='red')

show(p1+p2+p3)

Solution.

f(y)=2*y-y^2

g(y)=pi*f^2

print integral(g(y), y, 0, 2)

plot(f(y), y, 0, 2)

Solution.

Solution.

var('y')

f(y)=pi*(y^2-y^4)

print integral(f(y), y, 0, 1)

plot(f(y),0,1)

Solution.

var('x')

f(x)=x^(1/3)

g(x)=pi*f^2

print integral(g(x), x, 0, 1)

Solution.

var('x, y')

f(x)=pi*(1-x^2)^2-pi*(1-x)^2

print integral(f(x), x, 0, 1)

p1=plot(x, x, 0, 1)

p2=plot(x^2, x, 0, 1)

p3=parametric_plot((x,1), (x, 0, 1), color='red')

show(p1+p2+p3)

var('x, y, z')

p1=implicit_plot3d( y^2+z^2==(1-x)^2 ,

(x,0, 1), (y,-1, 1), (z,-1, 1), opacity=0.5, color="red")

p2=implicit_plot3d( y^2+z^2==(1-x^2)^2 ,(x,0,1), (y,-1, 1), (z,-1, 1), opacity=0.5, color="blue")

show(p1+p2)

Solution.

Solution.

var('x, y')

f(x)=1/x^2

p1=plot(f(x), x, 0.5, 4)

p2=parametric_plot((1,y), (y, -2, 4), color='red')

p3=parametric_plot((3,y), (y, -2, 4), color='red')

p4=parametric_plot((x,-1), (y, -2, 4), color='red')

p5=parametric_plot((x,0), (x, -1, 4), color='red')

show(p1+p2+p3+p4+p5)

cover1=(1,x)

cover2=(3,x)

sur1=revolution_plot3d(f(x),(x,1,3),axis=(0,-1),opacity=0.8,show_curve=True,parallel_axis='x')

sur2=revolution_plot3d(0,(x,1,3),axis=(0,-1),opacity=0.8,show_curve=True,parallel_axis='x')

sur3=revolution_plot3d(cover1,(x,0,f(1)),axis=(0,-1),show_curve=True,opacity=0.8,parallel_axis='x')

sur4=revolution_plot3d(cover2,(x,0,f(3)),axis=(0,-1),show_curve=True,opacity=0.8,parallel_axis='x')

(sur1+sur2+sur3+sur4).show()

A(x)=((f(x)-(-1))^2-(0-(-1))^2)*pi

integral(A(x),x,1,3)

http://matrix.skku.ac.kr/cal-lab/cal-6-2-14.html

Solution.

var('x, y')

f=x-y^2

p1=implicit_plot(f, (x, 0, 1.5), (y, -1.5, 1.5))

p2=parametric_plot((1,y), (y, -1.5, 1.5), color='red')

show(p1+p2)

g1=(x^2,x)

g2=(x^2,-x)

sur1=revolution_plot3d(g1,(x,0,1),axis=(1,0),show_curve=True,opacity=0.5)

sur2=revolution_plot3d(g2,(x,0,1),axis=(1,0),show_curve=True,opacity=0.5)

(sur1+sur2).show()

A(y)=(1-y^2)^2*pi

integral(A(y),y,0,1)

Solution.

var('x, y')

f(y)=pi*(1-y^2)^2-pi*(1-y)^2

print integral(f(y), y, 0, 1)

g(x)=x

p1=plot(g(x), x, -1, 2)

p2=plot(sqrt(x), x, 0, 2)

p3=parametric_plot((1,y), (y, -1, 2), color='red')

show(p1+p2+p3)

var('x, y, z')

p1=implicit_plot3d( (1-x)^2+z^2==y ,   (x,0, 2), (y,0, 1), (z,-1, 1), opacity=0.5, color="red")

p2=implicit_plot3d( (1-x)^2+z^2==y^2 ,(x,0, 2), (y,0, 1), (z,-1, 1), opacity=0.5, color="blue")

show(p1+p2)

Solution.

var('x, y')

f(y)=pi*y-pi*(y^2)^2

print integral(f(y), y, 0, 1)

g=x-y^2

p1=plot(x^2, x, -1, 1.5)

p2=implicit_plot(g, (x, -1, 2), (y, -1, 2))

p3=parametric_plot((0,y), (y, -1, 3), color='red')

show(p1+p2+p3)

Solution.

var('x, y')

f(x)=x

p1=plot(f(x), x, 0, 4)

p2=parametric_plot((1,y), (y, 0, 4), color='red')

p3=parametric_plot((3,y), (y, 0, 4), color='red')

p4=parametric_plot((x,0), (x, 0, 4), color='red')

show(p1+p2+p3+p4)

cover1=(1,y)

cover2=(3,y)

cover3=(x,0)

sur1=revolution_plot3d(f(x),(x,1,3),axis=(1,0),opacity=0.7,show_curve=True,parallel_axis='z')

sur2=revolution_plot3d(cover1,(y,0,f(1)),axis=(1,0),opacity=0.7,show_curve=True,parallel_axis='z')

sur3=revolution_plot3d(cover2,(y,0,f(3)),axis=(1,0),opacity=0.7,show_curve=True,parallel_axis='z')

sur4=revolution_plot3d(cover3,(x,1,3),axis=(1,0),opacity=0.7,show_curve=True,parallel_axis='z')

(sur1+sur2+sur3+sur4).show()

print pi*integral((3-1)^2-(y-1)^2, y, 1, 3)+4*pi

18-23. Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

Solution.

Solution.

Solution.

var('x, y')

f(x)=pi-pi*(1-cos(x))^2

g(X)=pi*(1-cos(x))^2-pi

print integral(f(x), x, 0, pi/2)+integral(g(x), x, pi/2, pi)

p1=plot(cos(x), 0, pi)

p2=parametric_plot((x,1), (x, -1, 4), color='red')

p3=parametric_plot((x,0), (x, -1, 4), color='green')

p4=parametric_plot((pi,y), (y, -1, 1), color='red')

show(p1+p2+p3+p4)

Solution.

Solution.

var('y')

f(y)= 3^2 - (1+sqrt(1+y^2))^2

pi*integral(f(y), y, -sqrt(3), sqrt(3))

http://matrix.skku.ac.kr/cal-lab/cal-6-2-23.html

Solution.

var('x, y')

f(y)=pi*(2-(y-1)^2+5)^2-pi*(1-y+5)^2

print integral(f(y), y, 0, 3)

p1=plot(-x+1, x, -6, 3)

g=(y-1)^2-2+x

p2=implicit_plot(g, (x, -6, 3), (y, -4, 4))

p3=parametric_plot((-5,y), (y, -4, 6), color='red')

show(p1+p2+p3)

24-27. Describe the solid of revolution whose volume is represented by the integral.

24.

Solution. ;

var('x, y, z')

implicit_plot3d(y^2 + z^2 == cos(x) , (x,0, pi/2), (y,-2, 2), (z,-2, 2), opacity=0.5, color="red")

integral (pi*cos(x), x, 0, pi/2)

25.

Solution. ;

var('x')

f(x)=x

p1=plot(f(x), x, 0, 6, ymin=-1, ymax= 6)

p2 = parametric_plot((x,2), (x, 0, 6), color='green')

p3 = parametric_plot((x,5), (x, 0, 6), color='red')

show(p1+p2+p3)

cover1=(x,5)

cover2=(x,2)

cover3=(0,x)

sur1=revolution_plot3d(f(x),(x,2,5),show_curve=True,opacity=0.7)

sur2=revolution_plot3d(cover1,(x,0,5),show_curve=True,opacity=0.7)

sur3=revolution_plot3d(cover2,(x,0,2),show_curve=True,opacity=0.7)

sur4=revolution_plot3d(cover3,(x,2,5),show_curve=True,opacity=0.7)

(sur1+sur2+sur3+sur4).show()

26.

Solution. ;

var('x, y, z')

expand (integral (pi*(x^2-x^4), x, 0, 1))

implicit_plot3d(x^2 + z^2 == y^2-y^4 , (x,-1, 1), (y,0, 1), (z,-1, 1), opacity=0.5, color="red")

27.

http://matrix.skku.ac.kr/cal-lab/cal-6-2-27.html

var('x')

f(x)=1-cos(x)

p1=plot(f(x), x, 0, pi/2, ymin=-3, ymax= 5)

p2 = parametric_plot((x,1), (x, 0, pi/2), color='green')

show(p1+p2)

cover1=(x,1)

cover2=(0,x)

sur1=revolution_plot3d(f(x),(x,0,pi/2),show_curve="True",opacity=0.7,parallel_axis='x')

sur2=revolution_plot3d(cover1,(x,0,pi/2),show_curve="True",opacity=0.7,parallel_axis='x')

sur3=revolution_plot3d(cover2,(x,0,1),show_curve="True",opacity=0.7,parallel_axis='x')

(sur1+sur2+sur3).show()

28-32. Find the volume of the described solid .

28. A right circular cone with height and base radius .

Solution.

29. A frustum of a right circular cone with height , lower base radius , and top radius .

Solution.

CAS 30. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 6cm, 8cm and 10cm.

Solution. ,

31. The base of S is an elliptical region with boundary curve . Cross-sections perpendicular to the -axis are isosceles right triangles with the hypotenuse in the base.

Solution.

32. A bowl is shaped like a hemisphere with diameter 40 cm. A ball with diameter 20 cm is placed in the bowl and water is poured into the bowl to a depth of centimeters. Find the volume of water in the bowl.

Solution. .

Solution

var('x, y')

x=sqrt(-y^2+4)+2

pi*integral(x, y, 0, 2)

34-37. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

34. , ; about the -axis

Solution.

CAS 35. , ; about -axis

Solution.

Solution.

var('x, y')

f(y)=pi*(5-y^2)^2-pi

print integral(f(y), y, -2, 2)

p1=parametric_plot((0,y), (y, -3, 3), color='red')

p2=parametric_plot((-1,y), (y, -3, 3), color='blue')

f=4-y^2-x

832/15*pip3832/15*pi=implicit_plot(f, (x, -2, 5), (y, -3, 3))

show(p1+p2+p3)

http://matrix.skku.ac.kr/cal-lab/cal-6-2-37.html

Solution.

var('x')

f(x)= -3*x^4 -36*x^2 + 192

g(x)= 2*pi*f

print integral(g(x), x, 0, 2)

plot(f(x), x, 0, 2)

CAS 38. Find the volume of the solids obtained by rotating the region bounded by the curves and about the following lines:

(a) the -axis

(b) the -axis

(c)

Solution. (a)

var('x, y')

f(x)= x^2 -4*x^4

g(x)= pi*f

print integral(g(x), x, 0, 1/2)

plot(f(x), x, 0, 1/2)

(b)

(c)

39. Let be the region in the first quadrant bounded by the curves and . Calculate the following quantities.

(a) The area of .

(b) The volume obtained by rotating about the -axis.

(c) The volume obtained by rotating about the -axis.

which is

var('x, y')

f(x)= x -2*x^3 -x^2

print integral(f(x), x, 0, 1/2)

plot(f(x), x, 0, 1/2)        #   5/96

(b)

(c)

6.3 Volumes by Cylindrical Shells

Suppose, for example, is the region bounded by , and and is true solid obtained by revolving about the -axis (see Figure 2).

When we use the formula  for the volume, we must solve equation for to get an equation of the form . That task may be not easy, so in this section we are going to present another approach, one that does not require solving the equation for .

Method of cylindrical shells: Figure 1 shows a cylindrical shell with inner radius , outer radius , and height . Its volume is calculated by subtracting the volume of the inner cylinder from the volume of the outer cylinder:

.

Figure 1

Let the thickness of the shell be and the average radius of the shell be . Then the formula for the volume of a cylindrical shell is

[1]

(circumference)(height)(thickness of shell)

Now let be the solid obtained by rotating about the -axis the region bounded by (where ), , , and , where . (See Figure 2.)

Figure 2

Divide the interval into subintervals of equal width and let be the midpoint of the th subinterval. If the rectangle with base and height is rotated about the -axis, then the result is a cylindrical shell with average radius , height , and thickness (see Figure 3), so by Formula [1] its volume is

.

Therefore, an approximation to the volume of is given by the summation of the volumes of these shells:

As we have

.

Thus,

Figure 3

The volume of the solid in Figure 3, obtained by rotating about the -axis the region under the curve from to , is

[2]                         where    .

Remember Formula [2] by imagining a typical shell, cut and flattened as in Figure 4, with radius , circumference , height and thickness or :

(circumference)(height)(thickness of shell).

Figure 4

Draw a diagram to help identify the radius and height of a shell.

EXAMPLE 1

The region between the line and curve rotated about the -axis to form a solid of revolution . Find the volume of the solid of revolution.

Figure 5

Solution. We use the cylindrical shell method because is the dependent variable. We see that curve crosses the -axis at and , and sketch the region in Figure 5. The volume is

.

show(plot(3*x-x^2, x, 0, 3, ticks=[[],[]], fill='axis', fillcolor='wheat') + plot(3*x-x^2, 1.9, 2, fill='axis'))

revolution_plot3d(3*x-x^2 , (x, 0, 3), show_curve= True, opacity=0.7).show(aspect_ratio=(1,1,1), parallel_axis='y')

EXAMPLE 2

The region between the curves and is rotated about the -axis. Find the volume of the solid of revolution.

Figure 6

Solution. We make a sketch in Figure 6 and find that the curves cross at and . We take for the independent variable and use the Cylindrical Shell Method.

Some regions are more easily described by taking as the independent variable, so that is the region between and for . The volumes of the solids of revolution are then computed by integrating with respect to . Often we have a choice of either or as the independent variable.

var('u')

sur1=revolution_plot3d(sqrt(u), (u, 0, 1), opacity=0.5, show_curve=True, parallel_axis='z')

sur2=revolution_plot3d(u, (u, 0, 1), opacity=0.5, show_curve=True, parallel_axis='z')

(sur1+sur2).show(aspect_ratio=(1, 1, 1))

plot(sqrt(x), x, 0, 1, ticks=[[],[]], fill=x,fillcolor='wheat') + plot(sqrt(x), x, 0.5, 0.55, ticks=[[],[]], fill=x,) + plot(x, 0, 1)

EXAMPLE 3

Determine the volume of the solid obtained by rotating about the -axis the region under the curve from to using cylindrical shells.

Solution. Rewrite the curve as in Figure 7. For rotation about the -axis we see that a typical shell has radius , circumference , and height . So the volume is

Figure 7

EXAMPLE 4

Compute the volume of the solid obtained by rotating the region bounded by and about the line .

Solution. The region and a cylindrical shell formed by rotation about the line are shown in Figure 8. It has radius , circumference , and height .

Figure 8

The volume of the given solid is

CAS EXAMPLE 5

Find the volume generated by rotating the region bounded by the given curves about the –axis using the method of cylindrical shells.

Figure 9                                       Figure 10

var('x, y')

p1 = plot(-x+2*sin(x), (x,0,pi/2), rgbcolor='red');

p2 = parametric_plot((x,0),(x,0,pi/2), rgbcolor='blue');

p3 = parametric_plot((0,y),(y,-1,1), rgbcolor='blue');

p4 = parametric_plot((pi/3,y),(y,-1,1), rgbcolor='blue');

show(p1+p2+p3+p4, aspect_ratio=1)

print integral(2*pi*x*(-x+2*sin(x)), x, 0, pi/3)

Answer : -2/81*(27*pi + pi^3 - 81*sqrt(3))*pi

Figure 11

plot(-x+2*sin(x),x,0,pi/3,ticks=[[],[]],fill='axis',fillcolor='wheat')+plot(-x+2*sin(x),x,0.5,0.55,fill='axis',ticks=[[],[]])+plot(-x+2*sin(x),0,pi/3)

revolution_plot3d(-x+2*sin(x),(x,0,pi/3),show_curve=False,opacity=0.7).show(aspect_ratio=(1,1,1),parallel_axis='y')

6.3 EXERCISES (Volumes by Cylindrical Shells)

CAS 1-10. Find the volume generated by rotating the region bounded by the given curves about the -axis using the method of cylindrical shells.

[You may refer to other open resources in http://math1.skku.ac.kr/ ]

Solution.

var('x, y')

p1=plot(2/x, (x, 0, 10),rgbcolor=(1, 0, 0));

p2=plot(0, (y, -10, 10),rgbcolor=(0, 1, 0));

show(p1+p2, aspect_ratio=1, xmin=-1, ymax=10)

print integral(2*x*pi*2/x, x, 2, 3)

Solution.

var('x,y')

p1=plot(-x^2+4 ,(x, -3, 3), rgbcolor=(1, 0, 0));

p2=plot(-2*x+4, (x, -3, 3), rgbcolor=(0, 1, 0));

show(p1+p2, aspect_ratio=1)

solve(-x^2+4==-2*x+4, x)

integral(2*pi*x(-x^2+4-(-2*x+4)), x, 0, 2)

Solution.

var('x,y')

p1=plot(3*(x-3)^2, (x, 0, 5), color="red");

p2=plot(x^2-6*x+11, (x, 0, 5), color="green");

show(p1+p2, aspect_ratio=1)

solve(3*(x-3)^2==x^2-6*x+11, x)

integral(2*pi*x(x^2-6*x+11-(3*(x-3)^2)), x, 2, 4)

Solution.

var('x,y')

p1=plot(1+2*y^2, (y, -3, 3), color="red");show(p1, aspect_ratio=1)

integral(2*pi*y*(1+2*y^2), y, 1, 2)

Solution.

var('x, y')

f(x)=2*x^2

p = plot(f(x),(x, 1, 2),color ='red')

show(p)

integral(2*pi*x*(2*x^2),x,1,2)

Solution.

var('x,y')

f(x)=2*x-x^2

g(x)=4*x-2*x^2

p1=plot(f(x),(x,-1,3),rgbcolor=(1,0,0));

p2=plot(g(x),(x,-1,3),rgbcolor=(0,1,0));

show(p1+p2,aspect_ratio=1)

solve(f(x)==g(x),x)

integral(2*pi*(x+1)*(g(x)-f(x)),x,0,2)

Solution.

var('x,y')

f(x)=ln(x)

g(x)=0

p1=plot(f(x),(x,0,3),rgbcolor=(1,0,0));

p2=plot(g(x),(x,0,3),rgbcolor=(0,1,0));

show(p1+p2,aspect_ratio=1)

integral(2*pi*x*ln(x),x,1,3)

Answer : (9 ln3 - 4)*pi

Solution.

var('x,y')

assume(x>=0)

f(x)=x^4-1

g(x)=cos(pi/2)

p1=plot(f(x),(x,-1,1),rgbcolor=(1,0,0));

p2=plot(g(x),(x,-1,1),rgbcolor=(0,1,0));

show(p1+p2,aspect_ratio=1)

integral(2*pi*(2+x)*(g(x)-f(x)),x,-1,1)

Solution.

var('x')f(x)=x^2-5*x+6

g(x)=0

p1=plot(f(x),(x,0,4),rgbcolor=(1,0,0));

p2=plot(g(x),(x,0,4),rgbcolor=(0,1,0));

show(p1+p2,aspect_ratio=1)

solve(f(x)==g(x),x)

integral(2*pi*x*abs(f(x)),x,2,3)

Solution.

var('x')

f(x)=2

g(x)=x^2-4*x+2

p1=plot(f(x),(x,0,4),rgbcolor=(1,0,0));

p2=plot(g(x),(x,0,4),rgbcolor=(0,1,0));

show(p1+p2,aspect_ratio=1)

solve(f(x)==g(x),x)

integral(2*pi*(x+1)*(f(x)-g(x)),x,0,4)

6.4 Work

In this section we apply the definite integral to find the amount of work that is done by a force moving an object. Work is the total amount of effort required to perform a task. You may think of a force as describing a push or pull on an object. Consider an object moving along a straight line with position function . Then by Newton’s Second Law of Motion the force on the object (in the same direction) is

[1]                                   .

Thus, a force of 1newton (N = ) acting on a mass of 1kg produces an acceleration of .

In a first course in Physics you will typically look at the work done by a constant force, . If the (net) force acting on an object is constant, then the object's acceleration is constant (from [1] with constant mass ). In this case the work done is defined to be the product of the force and the distance that the object moves:

(workforcedistance)

The unit for is a newton-meter, which is called a joule . Assume that the acceleration due to gravity is .

EXAMPLE 1

Find the work done in lifting a book off the floor to put it on a desk that is high.

Solution. From [1] the force exerted is equal and opposite to that exerted by gravity , so

.

Then from [2] the work done is

.

Variable Force

However, most forces are not constant and will depend the exact position where the force is acting.  So, let us assume that the force at any is given by .  Then the work done in moving the object from to is

[3]

To see a justification of this formula see the Proof of Various Integral Properties1) by Paul Dawkins. Notice that if the force is constant we get the same formula for a constant force where is simply the distance moved, . So, let us look at examples with non-constant forces.

EXAMPLE 2

Find the work done in moving a particle from to when it is located at a distance meters from the origin, and a force of Newtons acts on it.

Solution.

The work done is (Joule).

Hooke’s Law

The force, , required to maintain a spring stretched units beyond its natural length is proportional to , so where is a positive constant (called the spring constant) provided that is not too large. (See Figure 1.)

frictionless surface

(a)Natural position of spring          (b) Stretched position of spring

Figure 1 Hooke’s Law

EXAMPLE 3

A spring has been stretched from its natural length of 20cm to a length of 30cm by a force of 40newton(N). Find the work done in stretching the spring from 40cm to 50cm.

Solution. This example will require Hooke’s Law to determine the force.  Hooke’s Law tells us that the force required to stretch a spring a distance of x meters from its natural length is, where is called the spring constant. The first thing that we need to do is determine the spring constant for this spring. We can do that using the initial information. A force of 40 N is required to stretch the spring 30 cm – 20 cm = 10 cm = 0.10 m from its natural length. Using Hooke’s Law we have, . Thus

.

Now let us find the work done in stretching the spring from 40 cm to 50 cm. First we need to convert these distances from the natural length in centimeters, to meters, doing this gives values of 0.2 (=0.40-0.20) m and 0.3 (=0.50-0.20) m. The work is then,

EXAMPLE 4

We have a cable that weighs 2lbs attached to a bucket filled with coal that weighs 700lbs. The bucket is initially at the bottom of a 500ft mine shaft. Determine the amount of work required to lift the bucket all the way up the shaft.

Solution. Before answering we first need to determine the force. In this case the force will be the weight of the bucket and cable at any point in the shaft. To determine a formula for this first we will need to set a convention for , so for this problem, we will set to be the amount of cable that has been pulled up. So at the bottom of the shaft , and at the top of the shaft . Also at any point in the shaft there is feet of cable still in the shaft.

Then the force then for any is nothing more than the weight of the cable and bucket at that point. This is,

weight of cable + weight of bucket/coal

We can now answer the question. Then the total work done is

(ft-lb).

EXAMPLE 5

A tank has the shape of an inverted circular cone with height 15m and base radius 4meters  and is filled with water to a depth of 12 meters.  Determine the amount of work needed to pump all of the water to the top of the tank.  Assume that the density of the water is 1000.2)

Solution. In this case we cannot determine a force function, simply. So, we are going to need to approach this from a different viewpoint. Let us first set to be the lower end of the tank/cone and to be the top of the tank/cone. With this definition of the -values we can now see that the water in the tank will correspond to the interval  [0, 12]. So, let us start by dividing [0, 12] into subintervals each of width and let be any point from the subinterval where . This divides the water into layers. The layer is approximated by a circular cylinder with radius and height . Now, for each subinterval we will approximate the water in the tank corresponding to that interval as a cylinder of radius and height . Here is a quick sketch of the tank. (See Figure 2.)

The red strip in the sketch represents the “cylinder” of water in the subinterval.  A quick application of similar triangles will allow us to relate to as follows. , so . Now, the mass, , of the volume of water, , for the subinterval is simply,

= density .

We know the density of the water (it was given) and because we are approximating the water in the subinterval as a cylinder we can easily approximate the volume, , and hence the mass of the water in the subinterval. The mass for the subinterval is approximately,

To raise this volume of water we need to overcome the force of gravity that is acting on the volume and that is, , where is the gravitational acceleration. The force to raise the volume of water in the subinterval is then approximately,

.

Next, in order to reach the top of the tank the water in the th subinterval will need travel approximately m to reach the top of the tank. Because the volume of the water in the subinterval is constant the force needed to raise the water through any distance is also a constant force. Therefore the work to move the volume of water in the subinterval to the top of the tank, i.e. raise it a distance of , is then approximately,

The total amount of work required to raise all the water to the top of the tank is then approximately the sum of each of the for . Or,

To get the actual amount of work we simply need to take . That is compute the following limit,

This limit of a summation should look familiar to you. It has probably been some time, but recalling the definition of the definite integral we can see that this is nothing more than the following definite integral,

J.

As we have seen in the previous example we sometimes need to compute “incremental” work and then use that to determine the actual integral we need to compute. This idea does arise on occasion and we should not forget it.

Figure 2

CAS EXAMPLE 6

A particle is moved along the -axis by a force that measures N at a point meters from the origin. Find the work done in moving the particle from the origin to a distance of 3m.

var('x')

f=3/(1+sin(x))

integral(f, x, 0, 3)

Answer : -6*cos(3)/(sin(3) + cos(3) + 1) - 6/(sin(3) + cos(3) + 1) + 6

* An illustration of Work in a vector field:

6.4 EXERCISES (Work)

Find the work done in pushing a car a distance of 10 while exerting a constant force of .

Solution. Since , .  (in Joules)

2. How much work is done by a weightlifter in raising a barbell from the floor to a height of meter?

Solution. Since ,   .  (in Newtons)   So .  (in Joules)

3. A particle is moved along the -axis by a force that measures pounds at a point feet from the origin. Find the work done in moving the particle from the origin to a distance of .

Solution. . Use a change of variable , so . Then

.  ()

var('x')

f(x)= 5/(2*x^2)

print integral(f, x, 1, 17)

4. Shown is the graph of a force function (in Newtons) that increases to its maximum value and then remains constant. How much work is done by the force in moving an object a distance of ?

Solution.

5. A force of is required to hold a spring stretched beyond its natural length. How much work is done in stretching it from its natural length to beyond its natural length?

Solution. Since and , .

By Hooke’s Law, .   So .

6. A spring has a natural length of . If a force is required to keep it stretched to a length of , how much work is required to stretch it from to ?

Solution. By Hooke’s Law, , so .

. (in Joules)

7. Suppose that of work is needed to stretch a spring from its natural length of to a length , how much work is needed to stretch it from to .

Solution. .

That is .  Hence . (in Joules)

8-14. Find the work done.

8. A heavy rope, long, weighs and hangs over the edge of a building high. How much work is done in pulling the rope to the top of the building?

Solution. . (in Joules)

9. A chain lying on the ground is long and its mass is . How much work is required to raise one end of the chain to a height of ?

Solution. , .

So, .  (in Joules)

10. A cable that weight is used to lift of coal up a mine shaft deep. Find the work done.

Solution. Since the cable weight is , the length of cable is .

So .  (in Joules)

11. A bucket that weighs and a rope of negligible weight are used to draw water from a well that is deep. The bucket is filled with of water and is pulled up at a rate of , but water leaks out of a hole in the bucket at a rate of . Find the work done in pulling the bucket to the top of the well.

Solution. Since , .  (in Joules)

12. An chain weighs and hangs from a ceiling. Find the work done in lifting the lower end of the chain to the ceiling so that it is level with the upper end.

Solution. , .

So .  (in Joules)

13. An aquarium long, side and deep is full of water. Find the work needed to pump half of the water out of the aquarium.

Solution. .  (in Joules)

14. A circular swimming pool has a diameter of , the sides are high and the depth of the water is . How much work is required to pump all of the water out over the side?

Solution. . (in Joules)

15. Newton's Law of Gravitation states that two bodies with masses and attract each other with a force

where is the distance between the bodies and is the gravitational constant. If one of the bodies is fixed, find the work needed to move the other from to .

Solution.   (in Joules)

16. Use Newton’s Law of Gravitation to compute the work required to launch a satellite vertically to center. Take the radius of Earth to be and

.

Solution. Use the result of problem 15, then . (in Joules)

6.5 Average Value of a Function

The average value of finitely many numbers is defined as

.

To compute the average value of a function , , divide the interval into equal subintervals, each with length . Take points in successive subintervals and calculate the average of the numbers :

Since , so and the average value becomes

.

As the limiting value is

.

Therefore, the average value of on the interval is defined as

EXAMPLE 1

Compute the average value of the function on the interval .

Solution.  Here and and . So

The following theorem (which is a consequence of the Mean Value Theorem for derivatives and the FTC) gives a number at which the value of is exactly equal to the average value of the function, that is, .

The Mean Value Theorem for Integrals

If is continuous on , then there exists a number in such that

.

Geometric interpretation : For positive functions , there exists a number such that the area of the rectangle with base and height is same as the area of the region under the graph of from to . (See Figure 1.)

Figure 1

EXAMPLE 2

Verify the Mean Value Theorem for Integrals for on the interval .

Solution.  Since is continuous on the interval , by the Mean Value Theorem for Integrals there exists a number in such that

.

But from Example 1 above we know that so the value of satisfies

.

Therefore,

.  Thus, in this case there exists two numbers in the interval for which the condition of the Mean Value Theorem for Integrals is valid. Examples 1 and 2 are illustrated by Figure 2.

Figure 2

Average and Instantaneous Velocity

For an object moving in a straight line with position at time , the average velocity from time to time is the average rate of change of position with respect to time:                        (Average velocity)

Average rate of change of position

The instantaneous velocity at time is

(Instantaneous velocity)

In other words, instantaneous velocity is the derivative of position with respect to time.

EXAMPLE 3

A car travels with velocity , where is time. Between times and find the average velocity with respect to time.

Solution.  Average velocity with respect to time is

which agrees with . (See Figure 3.)

Figure 3

CAS EXAMPLE 4

Find the average value of the function on the given interval.

Solution.

var('x, y')

f=sin(sqrt(x))

inc=2-1

fave=integral(f, x, 1, 2)/inc; fave

p1 = plot(f, (x, 1, 2), rgbcolor='red')

p2 = parametric_plot((x, fave), (x, 1, 2), rgbcolor='blue')

p3 = parametric_plot((1, y), (y, 0.5, 1), rgbcolor='green')

p4 = parametric_plot((2, y), (y, 0.5, 1), rgbcolor='green')

show(p1+p2+p3+p4, aspect_ratio=1)

Figure 4

CAS EXAMPLE 5

Verify the Mean Value Theorem for in the interval .

http://matrix.skku.ac.kr/cal-lab/cal-6-5-exm-5.html

Solution.  By the Mean Value Theorem for Integrals, we have a number 0.21 in such that as shown in the following graph.

f(x)=exp(-x^2)*sin(x)

a=0

b=2

f.plot(a, b, legend_label="$e^{-x^2}\sin(x)$").show(figsize=3)

avg=f.integrate(x, a, b)/(b-a).n()

avg

s=numerical_integral(f(x)/(b-a), a, b);s

avg=s[0];avg             # 0.21

(plot(f, a, b, fill=True)+plot(avg, a, b)).show(figsize=3)

solve(f(x)==avg, x)

c1=find_root(f(x)==avg, 0, 1);c1

c2=find_root(f(x)==avg, 1, 2);c2

(plot(f, a, b)+plot(avg, a, b, fill=True) + line2d([(2, 0), (2, f(c1))])).show(figsize=3)

Figure 5                                 Figure 6                                Figure 7

6.5 EXERCISES (Average Value of a Function)

1-7. Find the average value of the function on the given interval.

1. ,

Solution. .

var('x')

integral(x^3,x,-1,1)/(1-(-1))

2. ,

Solution.

var('x')

fave(x)= (1/2)*ln(x)

print integral(fave(x), x, 2, 4)

plot(fave(x), x, 2, 4)    # 3*log(2) - 1

3. ,

Solution.

var('x')

Tave(x)= (2/pi) * (sin(x))

print integral(Tave(x), x, 0, pi/2)

plot(Tave(x), x, 0, pi/2)

4. ,

Solution.

To calculate this integral, use a change of variable , then . So,

5. ,

Solution.

var('x')

integral(2^(x)*ln(2),x,0,5)/(5-0)

6. ,

Solution.

7. ,

Solution. .

Use a change of variable , then .

So

var('x')

integral(2/(1+2*x)^2,x,1,6)/(6-1)

8-10. (a) Find the average value of on the given interval.

(b) Find such that .

(c) Sketch the graph of and a rectangle whose area is the same as the area under the graph of .

8. ,

Solution. (a)

(b)

var('x')

f(x)=(2*(x-3)^2)

plot(f(x),x,2,5

9. ,

Solution. (a)

(b)

10. , .

Solution. (a) .

Use change of variable to , then .

So

Also change of variable to ,  then .

var('x')

integral(x/((1+x^2)^2),x,0,2)/(2-0)

(b)

(c) var('x')

f(x)=(x/(1+x^2)^2)

plot(f(x),x,0,2)

11. If is continuous and , show that takes on the value at least once on the interval .

Solution. Since is continuous and , there exists a constant in such that

.

12. Find the numbers such that the average value of on the interval is equal to .

Solution.

That is,

.

13. In a certain city the temperature (in ) hours after A.M. is modeled by the function

Find the average temperature during the period from A.M. to P.M.

Solution.

var('x')

Tave(x)= 1/12 * (45+12*cos(pi*x/12))

print integral(Tave(x), x, 0, 12)

plot(Tave(x), x, 0, 12)

14. If a cup of coffee has temperature in a room where the temperature is then, according to Newton’s Law of Cooling, the temperature of the coffee after minutes is .

What is the average temperature of the coffee during the first half hour?

Solution.

var('x')

Tave(x)= 1/30 * (25+80*e^(-x/40))

print integral(Tave(x), x, 0, 30)

plot(Tave(x), x, 0, 30)

Answer : 5/3*(15*e^(3/4) - 64)*e^(-3/4) + 320/3

15. The linear density in a rod long is , where is measured in meters from one end of the rod. Find the average density of the rod.

Solution. Let .

Then  .

Change of variable to , then . So

16. If denotes the average value of on the interval and , show that

Solution.

Calculus