Answer:[tex](1,1),\ \dfrac{1}{3}[/tex]
Step-by-step explanation:
Given
Equation of the curves are [tex]y=x^2,\ y^2=x[/tex]
The intersection of the curve is
[tex]\Rightarrow y^4-y=1\\\\\Rightarrow y(y^3-1)=0\\\\\Rightarrow y=0,1\\[/tex]
So, x coordinates are [tex]x=0,1[/tex]
points of intersection are[tex](0,0),(1,1)[/tex]
So, the area bounded between the curves
[tex]\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}[/tex]
The area bounded by them is [tex]\frac{1}{3}[/tex]
