Answer to Question #265026 in Calculus for Cherrie

Question #265026

(a)Find the area of the region enclosed by the parabola 𝑦 = 2π‘₯ βˆ’ π‘₯

2 and the π‘₯ axis.

(b) Find the value of π‘š so that the line 𝑦 = π‘šπ‘₯ divides the region in part (a) into two

regions of equal area.


1
Expert's answer
2021-11-15T08:43:57-0500

(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)=0\n\\\\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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