UNIT5

# UNIT5 - Unit 5 Area Between Curves and Volumes of...

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Unformatted text preview: Unit 5 Area Between Curves and Volumes of Revolution We have learned how to find the area beneath a positive function. In this section we will learn how to find more general areas, those bounded between two curves. For example, suppose we wish to find the area bounded by f ( x ) and g ( x ) between x = 1 and x = 3 as shown: f(x) g(x) 1 2 3 x y dx f(x) - g(x) Figure 1: Area bounded by two curves On the interval [1,3], we see that f ( x ) ≥ g ( x ), so the function ( f ( x ) − g ( x )) is positive on [1,3]. With this oservation in mind we can see that the problem of finding the area between two curves ( f ( x ) and g ( x ) here) can be reduced to one of finding the area under a positive function (( f ( x ) − g ( x )) here). So 1 we need not learn another technique for these problems, we simply reduce the problem as above and then apply our earlier method of sketching the graph, drawing a small strip and calculating its area, and then integrating to find the entire area. Example 1. Find the area bounded by the curves f ( x ) = x 2 + 1 and g ( x ) = x − 1 on the interval [1,3]. Solution: We first sketch the region involved, this yields figure 1 (previous page). Drawing the small strip we see that it has area: dA = ( f ( x ) − g ( x )) dx = [( x 2 + 1) − ( x − 1)] dx We may therefore represent the area in question by: A = R 3 1 [( x 2 − 1) − ( x − 1)] dx = R 3 1 [( x 2 − x + 2)] dx = h x 3 3 − x 2 2 + 2 x i 3 1 = [(9 − 9 2 + 6) − ( 1 3 − 1 2 + 2)] = 6 + 9 2 − 11 6 = 26 3 Example 2. Find the area bounded by the curves f ( x ) = 9 − x 2 and y = 5 Solution: We first sketch the region involved, this will tell us the interval involved ( by the intersection points of the two functions. Setting f ( x ) = 9 − x 2 = y = 5 we get x = ± 2 so we have the intersection points shown: 2 y=5 y=9-x 2 x 2 5 9 (2,5) (9-x ) - 5 2 dx 1 st Quadrant Figure 2: Area bounded by f ( x ) = 9 − x 2 , and y = 5 Drawing the small strip as shown we see that it has area: dA = [(9 − x 2 ) − (5)] dx = (4 − x 2 ) dx We may therefore represent the area in question by: A = R 2- 2 (4 − x 2 ) dx = h 4 x − x 3 3 i 2- 2 = [(8 − 8 3 ) − ( − 8 + 8 3 )] = 2(8) − 2 3 (8) = 32 3 Example 3. Find the area bounded by the curves f ( x ) = x 2 − 4 and g ( x ) = − x + 2 Solution: Once again we sketch the region involved and find the intersection points of the two functions in order to find the interval. Setting x 2 − 4 = − x + 2 3 we get...
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## This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.

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UNIT5 - Unit 5 Area Between Curves and Volumes of...

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