第4.5节 两条曲线之间的面积

4.5 AREA BETWEEN TWO CURVES

A region in the plane can often be represented as the region between two curves. For example, the unit circle is the region between the curves

shown in Figure 4.5.1. Consider two continuous functions f and g on [ a, b] such that f(x) ≤ g(x) for all x in [ a, b]. The region R, bounded by the curves

y = f(x), y = g(x), x = a, x= b,

is called the region between f(x) and g(x) from a to b. If both curves are above the x-axis as in Figure 4.5.2, the area of the region R can be found by subtracting the area below f from the area below g:

area of R = ___ g(x ) dx - _____ f(x) dx.

It is usually easier to work with a single integral and write

area of R = ___ ( g(x)- f(x)) dx.

Figure 4.5.1 Figure 4.5.2

In the general case shown in Figure 4.5.3, we may move the region R above the x-axis by adding a constant c to both f(x) and g(x) without changing the area, and the same formula holds:

area of R = ___ (g(x) + c)dx - ____ (f(x) + c) dx

=_______________(g(x) - f(x)) dx.

Figure 4.5.3

To sum up, we define the area between two curves as follows.

DEFINITION

If f and g are continuous and f(x) ≤ g(x) for a ≤ x ≤ b, then the area of the region R between f(x) and g(x) from a to b is defined as

______ (g(x) - f(x)) dx.

EXAMPLE 1 Find the area of the region between the curves y= ____ x - 1 and y=x

From x =1 to x=2. In Figure 4.5.4, we sketch the curves to check that

____ x -1 ≤ x for 1 ≤ x ≤ 2. Then

Figure 4.5.4

EXAMPLE 2 Find the area of the region bounded above by y = x+2 and below

by y=x².

Part of the problem is to find the limits of integration. First draw a

sketch (Figure 4.5.5). The curves intersect at two points, which can be

found by solving the equation x+2 = x² for x.

x² - (x+2) = 0, (x+1) (x+2) = 0,

x = -1 and x=2.

Then

Figure 4.5.5

EXAMPLE 3 Find the area of the region R bounded below by the line y = -1 and above by the curves y= x3 and y= 2-x. The region is shown in Figure 4.5.6.

Figure 4.5.6

This problem can be solved in three ways. Each solution illustrates a different trick which is useful in other area problems. The three corners of the region are:

(-1, -1), where y = x3 and y= -1 cross.

(3, -1), where y = 2 - x and y= -1 cross.

(1, 1), where y = x3 and y= 2-x cross.

Note that y =x3 and y=2 - x can cross at only one point because x3 is always increasing and 2 -x is always decreasing.

FIRST SOLUTION Break the region into the two parts shown in Figure 4.5.7:

R1 from x= -1 to x=1, and R2 from x=1 to x=3. Then

area of R = area of R1 + area of R2.

area of R1 = ___ x3 - (-1 ) dx = ___ x4 + x ]___ =2.

area of R2 = ___ (2-x) - (-1 ) dx = 3x - ___ x² ]___ =2.

area of R = 2+2=4.

Figure 4.5.7

SECOND SOLUTION Form the triangular region S between y= -1 and y=2 -x from -1 to 3. The region R is obtained by subtracting from S the region S1 show in Figure 4.5.8. Then

area of R = area of S - area of S1.

area of S = __ (2- x) - (-1) dx = 3x - __ x²]__ =8

area of S1 = __ (2- x) - x3 dx = 2x - __ x²- __x4]__ =4.

area of R = 8- 4 = 4.

Figure 4.5.8

THIRD SOLUTION Use y as the independent variable and x as the dependent variable. Write the boundary curves with x as a function of y.

y = 2 - x becomes x=2 -y.

y = x3 becomes x=y1/3.

The limits of integration are y= -1 and y=1 (see Figure 4.5.9). Then

A = ____ (2-y) - y1/3 dy = 2y - __ y² - __ y4/3]__ = 4.

As expected, all three solutions gave the same answer.

Figure 4.5.9

PROBLEMS FOR SECTION 4.5

In Problems 1-43 below, sketch the given curves and find the area of the region bounded by them.

1 f(x)= 0, g(x) = 5x - x², 0 ≤ x ≤ 4

2 f(x)= ____, g(x) = x², 1 ≤ x ≤ 4

3 f(x)= ______, g(x) = 1, -1 ≤ x ≤ 1

4 y= x-2, y=3x1/35 , 0 ≤ x ≤ 1

5 y=_____, y=_______, 0 ≤ x ≤ 4

6 y=_____, y=_______, -0 ≤ x ≤ 1

7 The x-axis and the curve y= -5+6x-x²

8 The x-axis and the curve y= 1-x4

9 The y-axis and the curve x= 25-y²

10 The y-axis and the curve x= y(8-y)

11 y =cos x, y=2cosx, -π/2≤x≤π/2

12 y =sin x cos x, y=1, 0≤x≤π/2

13 y = -sin x , y=sin x, 0≤x≤π

14 y = sin x , y=cos x, 0≤x≤π/4

15 y =sin x cos x, y=sin x, 0≤x≤π

16 y =sin ²x cos x, y=sin x cos x, 0≤x≤π/2

17 y=x, y=ex, 0≤x≤2

18 y=e-x, y=ex, 0≤x≤2

19 y= -e-x, y=ex, -1≤x≤1

20 y= xex², y=e, 0≤x≤1

21 y=_______, y=1, 0≤x≤2

22 y=_______, y=_______, 0≤x≤2

23 y=1/x, y=x, 0≤x≤2

24 y=________, y=______ , 0≤x≤1

25 f(x) =x3/2, g(x) =x2/3

26 y=x² -2x, y= x-2

27 y=x4 -2x², y= 2x²-2

28 y=x4 -1, y= x3 - x

29 y=x4/( x²+1), y= 1/(x² +1)

30 y=_________, y=________, 0≤x

31 y=2x², y=x² + 4

32 x=y², x=2 - y²

33 __ +__ = 1 and the x-and y-axes

34 x²y = 4, x² + y = 5 (first quadrant)

35 y=x______, y=2x

36 y=0, y=x3 +x +2, x = 2

37 y= 2x+ 4, y= 2 - 3x, y= -x

38 y=x² - 1, y=(x-1)², y=(x+1)²

39 y=___, y=1, y=10 - 2x

40 y=x-2, y=2 - x, y=_____

41 y= -x, y=______, y=3x - 2

42 y= -2, y=x3 +x, x + y= 3

43 y= x², y=2x-2, y=2x-3(first quadrant)

44 Find the area of the ellipse x²/a² +y²/b² =1. Use the fact that the unit circle has area π.

45 Sketch the four-sided region bounded by the lines y=1, y=x, y=2x, and

y=6-x and find its area.

46 Find the number c > 0 such that the region bounded by the curves y=x,y= -2x,

and x=c has area 6.

47 Find the number c such that the region bounded by the curves y=x² and

y=cx has area 9.

48 Find the number c >0 such that the region bounded by the curves y=x² and y=c has

area 36.

49 Find the number c >0 such that the region bounded by the curves y=x² and y=cx has

area 9.

50 Find the value of c between -1 and 2 such that the area of the region bounded by the

lines y= -x, y=2x, and y=1+cx is a minimum.

51 Find the value of c such that the line y=c bisects the region bounded by the curves

y= x² and y =1

52 Find the value of c such that the line y=cx bisects the region bounded by the x-axis

and the curve y= x- x²