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 Rcan 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 cto 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 ≤ xfor 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 = x²for 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 Rbounded 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 = x3and y= -1 cross.

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

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

 

Note that y =x3and y=2 - x can cross at only one point because x3is always increasing and 2 -xis always decreasing.

 

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

            R1from x= -1 to x=1, and R2from 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 Sbetween y= -1 and y=2 -x from -1 to 3. The region Ris obtained by subtracting from Sthe region S1show 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) - x3dx = 2x - __ x²- __x4]__=4.

                    area of R = 8- 4 = 4.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.5.8

 THIRD  SOLUTION  Use y as the independent variable and xas 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=chas area 6.

47  Find the number csuch that the region bounded by the curves y=x²and

    y=cxhas area 9.

48  Find the number c>0 such that the region bounded by the curves y=x²and y=c has

    area 36.

49  Find the number c >0 such that the region bounded by the curves y=x²and y=cx has

    area 9.

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

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

51  Find the value of csuch that the line y=cbisects the region bounded by the curves

     y= x²and y =1

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

    and the curve y= x- x²