Question

Find the area bounded between the curve $\mathrm{y}={\mathrm{x}}^2$ and $\mathrm{y}={\mathrm{x}}^3$.

• (A) $\frac{1}{2}$
• (B) $\frac{1}{12}$
• (C) $1$
• (D) $\frac{1}{24}$

Answer 1

The Correct Option is B

Explanation:
Given: $\mathrm{y}={\mathrm{x}}^3$ and $\mathrm{y}={\mathrm{x}}^2$Finding a point of intersection:$\Rightarrow {\mathrm{x}}^3-{\mathrm{x}}^2=0$$\Rightarrow {\mathrm{x}}^2(\mathrm{x}-1)=0$$\Rightarrow \mathrm{x}=0,1$Let us draw the graph of the curve $\mathrm{y}={\mathrm{x}}^2$ and $\mathrm{y}={\mathrm{x}}^3$

Let the required area be A.Using the formula of the area under the curve,$\mathrm{A}=|{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\mathrm{dx}|$$\Rightarrow \mathrm{A}=\left|{\int }_0^1({\mathrm{x}}^3-{\mathrm{x}}^2)\mathrm{dx}\right|$$=\left|{[\frac{x^4}{4}-\frac{{\mathrm{x}}^3}{3}]}_0^1\right|$Substitute the limit to evaluate the area:$\Rightarrow \mathrm{A}=|\frac{1}{4}-\frac{1}{3}-0+0|$$=\frac{1}{12}$Hence, the correct option is (B).