Question

Find the area bounded by the curve $y=2x-x^2$, and the line $y=x.$

Answer 1

Given curve is $y=2x-x^2$
$-y=x^2-2x$
$-y+1=x^2-2x+1$
$-(y-1)=(x-1{)}^2$
Which represents a downward parabola with vertex at $(1,1)$

Point of intersection of the parabola and the line $y=x$
Put $y=x$
$-(x-1)=(x-1{)}^2$
$-x+1=x^2-2x+1$
$\Rightarrow x^2-x=0$
$x(x-1)=0$
$\Rightarrow x=0,1$
$\therefore$ Points of intersections are (0, 0) and (1, 1).
$\therefore$ The area enclosed between the curve $y=2x-x^2$ and the line y = x
${\int }_0^1(2x-x^2-x)dx={\int }_0^1(x-x^2)dx={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1$
$=(\frac{1}{2}-\frac{1}{3})-(0-0)=\frac{1}{6}$ sq. unit.