The area of the region included between the parabola $y^{2}=x$ and the line $x+y=2$ in the first quardrant is
- $\frac{1}{6}$ sq. units
- $\frac{7}{6} \mathrm{sq} .$ units
- $\frac{1}{2} \mathrm{sq}$. units
- $\frac{2}{3}$ sq. units
Solution
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
$\begin{array}{l}
=\int_{0}^{1} \sqrt{x} d x+\int_{1}^{2}(2-x) d x \\
=\frac{2}{3}[x \sqrt{x}]_{0}^{1}+2[x]_{1}^{2}-\frac{1}{2}\left[x^{2}\right]_{1}^{2} \\
=\frac{2}{3}+2-\left(\frac{1}{2} \times 3\right)=\frac{2}{3}+2-\frac{3}{2}=\frac{4+12-9}{6}=\frac{7}{6}
\end{array}$Asked in: MHT CET 2020 (20 Oct Shift 1)