The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is
The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is
$\frac{1}{12}$
$\frac{1}{6}$
$\frac{1}{3}$
$1$
Solution
Given curves are $y=x^2$ and $y=x^3$.
Intersection point is $(1,1)$.
$\begin{aligned} \text { Area } & =\int_0^1\left(x^2-x^3\right) d x \\ & =\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 \\ & =\frac{1}{3}-\frac{1}{4}=\frac{1}{12} \text { sq unit }\end{aligned}$