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Find the areas bounded by the indicated curves.$$y=x-4 \sqrt{x}, y=0$$

$$A=\frac{128}{3}$$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 2

Areas by Integration

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we want to find the area bounded by the curves, Y equals x minus four. Route X and Y equals zero. This question is challenging our understanding of applications of integration and particularly testing our ability to find the area between two curves which is for vertical components equal to B Y U minus Y one X. And the horizontal components I equal interval C two d x two minus x one dy. If we sketch these curves out we can identify the area that's founded and then we can identify the bounds and relevant functions. Didn't do so. If we sketch this out we have four minus X. Going down to the origin and up between Y equals zero. In this curve we see the small mountain shape underneath the X axis which has intercept 16. Thus if we integrate from 0 to 16 4 minutes forward. Ex excellent. Forward X. We can solve that. We have a equals equals 0 to 16 X minus forward X equals expert over 28 30 X three half to 16 or area 128 33 where taking the magnitude or absolute value so that the area is not negative.

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