The area of the region included between the parabola \(y^{2}=x\) and the line \(x+y=2\) in the first quardrant is

Point of intersection of \(y^{2}=x\) and \(x+y=2\) is
\(
(2-x)^{2}=x \quad \Rightarrow x^{2}-4 x-x+4=0
\)
\(
x^{2}-5 x+4=0 \Rightarrow(x-4)(x-1)=0 \Rightarrow x=1,4
\)
But since we want area in \(1^{\text {st }}\) quadrant only, we take \(x=1\) \(\therefore y^{2}=1 \Rightarrow y=\pm 1 \Rightarrow y=1\) in \(1^{\text {st }}\) quadrant.
\(\therefore A \equiv(1,1)\) and \(P \equiv(1,0)\)
Point of intersection of \(x+y=2\) with \(X\) axis is \(B=(2,0)\)
Hence area required is
\(
\begin{array}{l}
=\int_{0}^{1} \sqrt{x} d x+\int_{1}^{2}(2-x) d x \\
=\frac{2}{3}[x \sqrt{x}]_{0}^{1}+2[x]_{1}^{2}-\frac{1}{2}\left[x^{2}\right]_{1}^{2} \\
=\frac{2}{3}+2-\left(\frac{1}{2} \times 3\right)=\frac{2}{3}+2-\frac{3}{2}=\frac{4+12-9}{6}=\frac{7}{6}
\end{array}
\)