Applications of Integration

6 Applications of Integration

6.1 Areas Between Curves

33

Areas Between Curves

Consider the region S that lies between two curves y = f (x) and y = g(x) and between the vertical lines x = a and x = b, where f and g are continuous functions and

f (x) g(x) for all x in [a, b]. (See Figure 1.)

Figure 1

S = {(x, y) | a x b, g(x) y ƒ(x)}

44


We divide S into n strips of equal width and then we approximate the i th strip by a rectangle with base x and height f (xi*) – g(xi*). (See Figure 2. If we like, we could take all of the sample points to be right endpoints, in which case xi* = xi.)

Figure 2

55


The Riemann sum

is therefore an approximation to what we intuitively think of

as the area of S.

This approximation appears to become better and better as

n . Therefore we define the area A of the region S as

the limiting value of the sum of the areas of these

approximating rectangles.

66


We recognize the limit in (1) as the definite integral of f – g.

Therefore we have the following formula for area.

Notice that in the special case where g(x) = 0, S is the region under the graph of f and our general definition of area (1) reduces.

77


In the case where both f and g are positive, you can see from Figure 3 why (2) is true:

A = [area under y = f (x)] – [area under y = g(x)]

Figure 3

88

Example 1

Find the area of the region bounded above by y = x2 + 1, bounded below by y = x, and bounded on the sides by x = 0 and x = 1.

Solution:

The region is shown in Figure 4. The upper boundary curve is y = x2 + 1 and the lower boundary curve is y = x.

Figure 4

99

Example 1 – Solution

So we use the area formula (2) with f (x) = x2 + 1, g(x) = x, a = 0, and b = 1:

cont’d

1010


If we are asked to find the area

between the curves y = f (x) and

y = g (x) where f (x) g (x) for

some values x of but g (x) f (x)

for other values of x, then we split

the given region S into several regions S1, S2 , . . . with areas A1, A2 , . . . as shown in Figure 9. We then define the area of the region S to be the sum of the areas of the smaller regions S1, S2 , . . . that is A = A1 + A2 + . . . Since

f (x) – g (x) when f (x) g (x)

| f (x) – g (x) | =

g (x) – f (x) when g (x) f (x)

Figure 9

1111


we have the following expression for A.

When evaluating the integral in (3), however, we must still split it into integrals corresponding to A1, A2,……

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Example 2

Find the area of the region bounded by the curves y = sin x, y = cos x, x = 0, and x = /2

Solution:

The points of intersection occur when sin x = cos x, that is, when x = /4 (since 0 x /2). The region is sketched in Figure 10. Observe that cos x sin x when 0 x /4 but sin x cos x when /4 x /2.

Figure 10

1313


Therefore the required area is

cont’d

1414


In this particular example we could have saved some work by noticing that the region is symmetric about x = /4 and so

cont’d

1515

Areas Between CurvesSome regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations x = f (y), x = g (y), y = c, and y = d, where f and g are continuous and f (y) g (y) for c y d (see Figure 11), then its area is

Figure 11

XL XR

1616

Find the area enclosed by the line y = x - 1 and the parabola y2 = 2x + 6.

Example 3

1717

By solving the two equations, we find that the points of intersection are (-1, -2) and (5, 4).

• We solve the equation of the parabola for x.

• From the figure, we notice that the left and right boundary curves are:

212 3

1L

R

x y

x y

Example 3 - solution

1818

We must integrate between the appropriate y-values, y = -2 and y = 4.

4

2

4 2122

4 2122

43 2

2

1 46 3

1 3

4

14

2 3 2

(64) 8 16 2 8 18

R LA x x dy

y y dy

y y dy

y yy

1919

In the example, we could have found the area by integrating with respect to x instead of y. However, the calculation is much more involved.

It would have meant splitting the

region in two and computing

the areas labeled A1 and A2.

• The method used in the example is much easier.

AREAS BETWEEN CURVES

2020

6.2 Volumes

2121

Volumes ,V = Ah

Figure 1(a)

Figure 1(c)Figure 1(b)

2222

Volumes by cross sections

2424

A(x) is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis.

2525

Volumes

4

dyyRb

a])([ 2

2626

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

The volume of the solid is generated by revolving the area between the graph y =f(x) of and the x-axis from x=a to x=b and around the x-axis is

The volume of the solid is generated by revolving the area between the graph x =g(y) of and the y-axis from y=c to x=d and around the y-axis is

dxxfVb

a 2)]([

dyygVd

c 2)]([

Solids of Revolution : The Disk Method

2929

Example 1Find the volume of the solid obtained by rotating about the 𝑥-axis the region under the curve 𝑦= ξ𝑥 from 0 to 1. Illustrate the definition of volume by sketching a typical approximating cylinder.

3030

Find the volume of the solid obtained by rotating the region

bounded by y = x3, y = 8, and x = 0 about the y-axis.

Example 2

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Example 2 – solutionIntegrate with respect to y.• Slicing at height y, we get a circular

disk with radius x, where

• So, the area of a cross-section

through y is:

• Since the solid lies between y = 0 and

y = 8, its volume is:

3x y

2 2 2/33( ) ( )A y x y y

8

0

8 2 3

0

853 35

0

( )

96

5

V A y dy

y dy

y

3232

Volumes

The Washers

3434

The region R enclosed by the curves y = x and y = x2 is rotated

about the x-axis. Find the volume of the resulting solid.

Example 3

The curves y = x and y = x2 intersect at the points (0, 0) and (1, 1).

The region between them, the solid of rotation, and cross-section perpendicular to the x-axis are shown.

3535

Thus, we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle:

Thus, we have:2 2 2

2 4

( ) ( )

( )

A x x x

x x

Example 3 -- solution

1

0

1 2 4

0

13 5

0

( )

( )

3 5

2

15

V A x dx

x x dx

x x

3636

Example 4

Find the volume of the solid obtained by rotating the region in Example 3 about the line y = 2.

Again, the cross-section is a washer.

the inner radius is 2 – x and

the outer radius is 2 – x2.

3737

6.3 Volumes by Cylindrical Shells

3838

Volumes by Cylindrical Shells

3939

Volumes by Cylindrical Shells• Its volume V is calculated by subtracting the volume V1 of the inner

cylinder from the volume V2 of the outer cylinder:

V = V2 – V1

= r22h – r1

2h = (r22 – r1

2)h

• = (r2 + r1)(r2 – r1)h

• = 2 h(r2 – r1)

V = [circumference] [height] [thickness]

4040


4141


4242

Example 1

Find the volume of the solid obtained by rotating

about the y-axis the region bounded by y = 2x2 – x3 and y = 0.

Solution:

From the sketch in Figure 6 we see that a typical shell has radius x, circumference 2x, and height f (x) = 2x2 – x3.

Figure 6

4343


So, by the shell method, the volume is

cont’d

4444

Find the volume of the solid obtained by rotating about the y-axis the region between y = x and y = x2.

We see that the shell has radius x, circumference 2πx, and height x - x2.

Example 2

4545

• Thus, the volume of the solid is:

1 2

0

1 2 3

0

13 4

0

2

2

23 4 6

V x x x dx

x x dx

x x

Example 2

4646

Example 3Find the volume of the solid obtained by rotating the region bounded by y = x - x2 and y = 0 about the line x = 2.

The figures show the region and a cylindrical shell formed by rotation about the line x = 2, which has radius 2 - x, circumference 2π(2 - x), and height x - x2.

4747

• So, the volume of the solid is:

0 2

1

0 3 2

1

143 2

0

2 2

2 3 2

24 2

V x x x dx

x x x dx

xx x

Example 3

4848

6.4 Average Value of a Function

4949

Average Value of a Function

It is easy to calculate the average value of finitely many numbers y1, y2, . . . , yn:

But how do we compute the average temperature during a day if infinitely many temperature readings are possible?

Figure 1 shows the graph of a temperature function T(t), where t is measured in hours and T in C, and a guess at the average temperature, Tave.

Figure 1

5050


We define the average value of f on the interval [a, b] as

5151

Example 1

Find the average value of the function f (x) = 1 + x2 on the interval [–1, 2].

Solution:

With a = –1 and b = 2 we have

5252


If T(t) is the temperature at time t, we might wonder if there is a specific time when the temperature is the same as the average temperature.

For the temperature function graphed in Figure 1, we see that there are two such times––just before noon and just before midnight.

In general, is there a number c at which the value of a function f is exactly equal to the average value of the function, that is, f (c) = fave?

Figure 1

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The following theorem says that this is true for continuous functions.

The Mean Value Theorem for Integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus.

5454

6.5 Arc Length

5555

Arc Length

dividing the interval up into n equal subintervals each of width

the point on the curve at each point is denoted by Pi.

We can then approximate the curve by a series of straight lines connecting the points.

the approximation gets better as we let n increase

Archimedes used the perimeters of inscribed polygons to approximate the circumference of a circle. For n=96, the approximation method gives 3.14103 as the circumference of the unit circle.

5656

the length of each of these line segments

and the length of the curve will then be approximately,

and we can get the exact length by taking n larger and larger.

Arc Length

11

limn

i in

i

L P P

PiPi 1

PiPL i

n

i

1

1

5757

If we let Δyi = yi – yi–1, then

Arc Length

2 21 1 1

2 2

( ) ( )

( ) ( )

i i i i i i

i

P P x x y y

x y

By the Mean Value Theorem we know that on the interval [xi–1, xi], we find that there is a number xi* between xi–1 and xi such that

that is,

*1 1( ) ( ) '( )( )i i i i if x f x f x x x

*'( )i iy f x x

http://tutorial.math.lamar.edu/Classes/CalcI/MeanValueTheorem.aspx

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Arc Length

Thus, we have:2 2

1

22 *

2* 2

2*

( ) ( )

( ) '( )

1 '( ) ( )

1 '( ) (since 0)

i i i

i

i

i

P P x y

x f x x

f x x

f x x x

Therefore,

11

2*

1

lim

lim 1 '( )

n

i ini

n

in

i

L P P

f x x

5959

Arc Length

If we use Leibniz notation for derivatives, we can write the

arc length formula as follows:

6060

Example 1

Find the length of the arc of the semicubical parabola

y2 = x

3 between the points (1, 1) and (4, 8). (See Figure 5.)

Figure 5

6161


For the top half of the curve we have

y = x3/2

So the arc length formula gives

If we substitute u = 1 + , then du = dx.

When x = 1, u = ; when x = 4, u = 10.

6262


Therefore

cont’d

6363

Arc Length

If a curve has the equation x = g (y), c y d, and g (y) is continuous, then by interchanging the roles of x and y in Formula 2 or Equation 3, we obtain the following formula for its length:

Arc Length formula:

6464

Find the length of the curve from 1≤y≤9.

Example 2

)3(3

1 yyx

6565

Example 2 -- solution

6666

6.6 Area of a Surface of Revolution

6767

Area of a Surface of Revolution

In this section we are going to look once again at solids of revolution. We first looked at themback the volume of the solid of revolution. In this section we wantto find the surface area of this region.

6868

Area of a Surface of RevolutionIf yi = f(xi), then the point Pi(xi, yi) lies on the curve.

The part of the surface between xi–1 and xi is approximated by taking the line segment Pi–1 Pi

and rotating it about the x-axis.

The result is a band with

slant height l = | Pi–1Pi |

and average radius r = ½(yi–1

+ yi). So, its surface area is:

112 | |

2i i

i i

y yP P

6969

2

1 1 '( *)i iP P f x x where xi* is some number in [xi–1, xi].

From the previous section on arc length:

When Δx is small, we have yi = f(xi) ≈ f(xi*) and yi–1 = f(xi–1) ≈ f(xi*), since f is continuous.

Therefore,

2* *112 2 ( ) 1 '( )

2i i

i i i i

y yP P f x f x x

7070

SURFACE AREAThus, an approximation to what we think of as the area of the complete surface of revolution is:

The approximation appears to become better as n → ∞.

2* *

1

2 ( ) 1 '( )n

i ii

f x f x x

2* *

1

2

lim 2 ( ) 1 '( )

2 ( ) 1 '( )

n

i in

i

b

a

f x f x x

f x f x dx

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Example 1

The curve y = , –1 x 1, is an arc of the circle

x2 + y2 = 4.

Find the area of the surface obtained by rotating this arc about the x-axis. (The surface is a portion of a sphere of radius 2. See Figure 6.)

Figure 6

7373


We have

and so, by Formula 5, the surface area is

7474


= 4 1 dx

= 4 (2)

= 8

cont’d

7575

SURFACE AREA• The arc of the parabola y = x2 from (1, 1)

to (2, 4) is rotated about the y-axis.

• Find the area of the resulting surface.

Example 2

7676

SURFACE AREA• Using y = x2 and dy/dx = 2x,

22

1

2 2

1

2

2 1

2 1 4

S x ds

dyx dx

dx

x x dx

E. g. 2—Solution 1

Substituting u = 1 + 4x2, we have du = 8x dx. Remembering to change the limits of integration, we have:

17 173 223 554 4

(17 17 5 5)6

S u du u

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SURFACE AREA• Using x = and dx/dy = ,

• we have the following solution.

y

E. g. 2—Solution 2

12 y

24

1

4

1

4

1

17

5(where 1 4 )

2 2 1

12 1

4

4 1

4

(17 17 5 5)6

u y

dxS xds x dy

dy

y dyy

y dy

udu

Applications of Integration

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