630
Chapter 6 | Applications of Integration
⎤ ⎦ | 1 2
⎡ ⎣ 2 x − x
1 2 (2− x ) dx =
2
A 2 = ∫
= 1 2 .
2
Adding these areas together, we obtain
5 6
1 2 =
A = A 1 + A 2 = 1
.
3 +
The area of the region is 5/6 units 2 .
Consider the region depicted in the following figure. Find the area of R .
6.4
Regions Defined with Respect to y In Example 6.4 , we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of y , instead of as functions of x ? Review Figure 6.7 . Note that the left graph, shown in red, is represented by the function y = f ( x ) = x 2 . We could just as easily solve this for x and represent the curve by the function x = v ( y ) = y . (Note that x =− y is also a valid representation of the function y = f ( x ) = x 2 as a function of y . However, based on the graph, it is clear we are interested in the positive square root.) Similarly, the right graph is represented by the function y = g ( x ) =2− x , but could just as easily be represented by the function x = u ( y ) =2− y . When the graphs are represented as functions of y , we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to y , we need to evaluate one integral only. Let’s develop a formula for this type of integration. Let u ( y ) and v ( y ) be continuous functions over an interval ⎡ ⎣ c , d ⎤ ⎦ such that u ( y ) ≥ v ( y ) for all y ∈ ⎡ ⎣ c , d ⎤ ⎦ . We want to find the area between the graphs of the functions, as shown in the following figure.
Figure 6.8 We can find the area between the graphs of two functions, u ( y ) and v ( y ).
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