# Answer in Calculus for Cherrie #265026

(a)Find the area of the region enclosed by the parabola = 2 −

2 and the axis.

(b) Find the value of so that the line = divides the region in part (a) into two

regions of equal area.

(a) “smallint_{0}^2 2x-x^2dx”

“=[x^2-frac{x^3}{3}]_{0}^2”

“=4-frac{8}{3}”

“=frac{4}{3}”

(b)

calculate half of the area found in part A

“=frac{4}{3}(frac{1}{2})n\=frac{2}{3}”

For the function y = mx, one of the points is the origin (0,0). we need to find the other point that intersects the curve. proceed as follows

“mx=2x-x^2\x^2+mx-2x=0\x^2+x(m-2)=0n\x=0 or x=2-m”

find x value in terms of m where curves intersect

“smallint_{0}^{2-m}(2x-x^2-mx)dx=frac{2}{3}”

“[x^2-frac{x^3}{3}-frac{mx^2}{2}]_{0}^{2-m}”

“(2-m)^2-frac{(2-m)^3}{3}-frac{m(2-m)^3}{2}=frac{2}{3}”

“(2-m)^2[frac{3}{3}-frac{2}{3}+frac{2m}{6}-frac{3m}{6}=frac{2}{3}”

“frac{1}{6}(2-m)^2[2-m]=frac{2}{3}”

“frac{1}{6}(2-m)^3=frac{2}{3}”

Evaluate and solve m

“(2-m)^3=frac{2}{3}(6)”

“(2-m)=sqrt[3]{4}”

“m=2-sqrt[3]{4}”

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