Question

This is an integration question.

Answer 1

Answer:$(1,1),\ \dfrac{1}{3}$

Step-by-step explanation:

Given

Equation of the curves are $y=x^2,\ y^2=x$

The intersection of the curve is

$\Rightarrow y^4-y=1\\\\\Rightarrow y(y^3-1)=0\\\\\Rightarrow y=0,1$

So, x coordinates are $x=0,1$

points of intersection are$(0,0),(1,1)$

So, the area bounded between the curves

$\Rightarrow I=\int_{0}^{1}\left ( \sqrt{x}-x^2\right )dx\\\\\Rightarrow I=\int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}x^2dx\\\\\Rightarrow I=\left ( \frac{2}{3}x^{\frac{3}{2}} \right )_0^1-\left ( \frac{1}{3}x^3 \right )_0^1\\\\\Rightarrow I=\frac{2}{3}\left ( 1-0 \right )-\frac{1}{3}\left ( 1^3-0 \right )\\\\\Rightarrow I=\frac{2}{3}-\frac{1}{3}\\\\\Rightarrow I=\frac{1}{3}$

The area bounded by them is $\frac{1}{3}$