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, y = g ( x ) = 2 − x, but could just as easily be represented by the function x = u ( y ) = 2 − y. (Note that x = − y x = − y is also a valid representation of the function y = f ( x ) = x 2 y = f ( x ) = x 2 as a function of y. ![]() We could just as easily solve this for x x and represent the curve by the function x = v ( y ) = y. Note that the left graph, shown in red, is represented by the function y = f ( x ) = x 2. What if we treat the curves as functions of y, y, instead of as functions of x ? x ? Review Figure 2.7. However, there is another approach that requires only one integral. In Example 2.4, we had to evaluate two separate integrals to calculate the area of the region. We want to find the area between the graphs of the functions, as shown in the following figure.Ĭonsider the region depicted in the following figure. Let f ( x ) f ( x ) and g ( x ) g ( x ) be continuous functions over an interval such that f ( x ) ≥ g ( x ) f ( x ) ≥ g ( x ) on. Last, we consider how to calculate the area between two curves that are functions of y. We then look at cases when the graphs of the functions cross. We start by finding the area between two curves that are functions of x, x, beginning with the simple case in which one function value is always greater than the other. In this section, we expand that idea to calculate the area of more complex regions. In Introduction to Integration, we developed the concept of the definite integral to calculate the area below a curve on a given interval.
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