Chapter 6 | Applications of Integration
629
| f ( x )− g ( x ) | = | sin x −cos x | = sin x −cos x .
Then
a b | f ( x )− g ( x ) | dx
A = ∫
0 π | sin x −cos x | dx = ∫ 0 π /4
π /4 π
= ∫
(cos x −sin x ) dx + ∫
(sin x −cos x ) dx
π /4 +[−cos x −sin x ] |
π /4 π
= [sin x +cos x ] | 0 = ( 2−1)+ ⎛
⎝ 1+ 2 ⎞
⎠ =2 2.
The area of the region is 2 2 units 2 .
6.3 If R is the region between the graphs of the functions f ( x ) = sin x and g ( x ) =cos x over the interval [ π /2, 2 π ], find the area of region R .
Example 6.4 Finding the Area of a Complex Region
Consider the region depicted in Figure 6.7 . Find the area of R .
Figure 6.7 Two integrals are required to calculate the area of this region.
Solution As with Example 6.3 , we need to divide the interval into two pieces. The graphs of the functions intersect at x =1 (set f ( x ) = g ( x ) and solve for x ), so we evaluate two separate integrals: one over the interval [0, 1] and one over the interval [1, 2]. Over the interval [0, 1], the region is bounded above by f ( x ) = x 2 and below by the x -axis, so we have A 1 = ∫ x 2 dx = x = 1 3 . Over the interval [1, 2], the region is bounded above by g ( x ) =2− x and below by the x -axis, so we have 0 1 3 3 | 0 1
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