Question

(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.

Answer 1

(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

$=\frac{4}{3}(\frac{1}{2}) \\=\frac{2}{3}$

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

$mx=2x-x^2\\x^2+mx-2x=0\\x^2+x(m-2)=0 \\x=0 \ or \ x=2-m$

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

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

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

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