The area bounded by the parabola \(y^2=x\) and the line \(x+y=2\) in the first quadrant is

The point of intersection of \(y^2=x\) and \(x+y=2\) is, \((2-x)^2=x \Rightarrow x^2-5 x+4=0 \Rightarrow(x-4)(x-1)=0\)
Let \(\mathrm{A}=(1,1)\) in first quadrant and \(\mathrm{B}=(4,-2)\) in fourth quadrant The line \(\mathrm{x}+\mathrm{y}=2\) cuts \(\mathrm{X}\) axis at \(\mathrm{P}(2,0)\)
Refer figure


Required area is shaded
\(
\begin{aligned}
& \therefore \mathrm{A}=\int_0^1 \sqrt{\mathrm{x}} \mathrm{dx}+\int_1^2(2-\mathrm{x}) \mathrm{dx} \\
& =\left[\frac{\mathrm{x}^{\frac{2}{3}}}{\left(\frac{3}{2}\right)}\right]_0^1+[2 \mathrm{x}]_1^2-\left[\frac{\mathrm{x}^2}{2}\right]_1^2 \\
& =\left[\left(\frac{2}{3}\right)(1)\right]+[2(2-1)]-\left[\left(\frac{4-1}{2}\right)\right]=\frac{2}{3}+2-\frac{3}{2} \\
& =\frac{7}{6} \text { sq. units }
\end{aligned}
\)