Answer to Question #265026 in Calculus for Cherrie

Question #265026

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

2 and the 𝑥 axis.

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

regions of equal area.


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

(a) 022xx2dx\smallint_{0}^2 2x-x^2dx


=[x2x33]02=[x^2-\frac{x^3}{3}]_{0}^2


=483=4-\frac{8}{3}


=43=\frac{4}{3}




(b)

calculate half of the area found in part A

=43(12)=23=\frac{4}{3}(\frac{1}{2}) \\=\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=2xx2x2+mx2x=0x2+x(m2)=0x=0 or x=2mmx=2x-x^2\\x^2+mx-2x=0\\x^2+x(m-2)=0 \\x=0 \ or \ x=2-m


find x value in terms of m where curves intersect

02m(2xx2mx)dx=23\smallint_{0}^{2-m}(2x-x^2-mx)dx=\frac{2}{3}


[x2x33mx22]02m[x^2-\frac{x^3}{3}-\frac{mx^2}{2}]_{0}^{2-m}


(2m)2(2m)33m(2m)32=23(2-m)^2-\frac{(2-m)^3}{3}-\frac{m(2-m)^3}{2}=\frac{2}{3}


(2m)2[3323+2m63m6=23(2-m)^2[\frac{3}{3}-\frac{2}{3}+\frac{2m}{6}-\frac{3m}{6}=\frac{2}{3}


16(2m)2[2m]=23\frac{1}{6}(2-m)^2[2-m]=\frac{2}{3}


16(2m)3=23\frac{1}{6}(2-m)^3=\frac{2}{3}

Evaluate and solve m

(2m)3=23(6)(2-m)^3=\frac{2}{3}(6)


(2m)=43(2-m)=\sqrt[3]{4}


m=243m=2-\sqrt[3]{4}



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