Question

The area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x-y$ plane is

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{6}$

Given, equation of straight line is
$y=x\left(1\right)$
And equation of parabola is
$y=x^2\left(2\right)$



From equation $\left(1\right)$ and $\left(2\right)$ we get
$x^2=x$
$\Rightarrow x^2-x=0$
$\therefore x=0.1$
$x=0$ and $x=1$
$\Rightarrow y=0$ and $y=1$
(Using vertical strip)
Enclosed Area $=\int \int \left(1\right)dydx$
$={\int }_{x=0}^1{\int }_{y=x^2}^x\left(1\right)dy.dx$
$={\int }_0^1(x-x^2)dx$
$={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$