Answer:
[tex]\frac{1}{6}[/tex]
Step-by-step explanation:
y = x .....(1)
[tex]y=\sqrt{x}[/tex] .... (2)
By solving equation (1) and equation (2)
[tex]x = \sqrt{x}[/tex]
[tex]\sqrt{x}\left ( \sqrt{x}-1 \right )=0[/tex]
[tex]\sqrt{x}=0[/tex] or [tex]\left ( \sqrt{x}-1 \right )=0[/tex]
x = 0, x = 1
y = 0, y = 1
A = [tex]\int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)[/tex]
A = [tex]\int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx[/tex]
A = [tex]\frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}[/tex]
A = [tex]\frac{2}{3}-\frac{1}{2}[/tex]
A = [tex]\frac{1}{6}[/tex]
