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Sketch the region enclosed by the given curves and find its area.$y=x^{2}, \quad y=4 x-x^{2}$

$\frac{8}{3}$

Calculus 2 / BC

Chapter 6

Applications of Integrals

Section 1

Areas Between Curves

Applications of Integration

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All right. So for this question, we have X squared and negative X squared plus four x You have X squared is equal to Native X squared plus or acts we have to find the intersection. Points just becomes X squared plus two x minus four X is equal to zero plus x squared, so this is equal to two X squared minus four. X is equal to zero with a factor of two. X to get X minus. Two is equal to zero, so we have actual equal to two on X is equal to zero. So therefore we can rewrite the integral as the integral from 2 4 0 2 2 0 2 0 0 to 2 of the ex queer minus negative X squared plus four X is equal to be integral from 4 0 to 2 of x two X squared minus four x This is to third X cubed minus two X squared months. We evaluate this from zero to and that that this is equal to eight over three

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