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

• Correct answer
  A

  $\frac{7}{6}$ aq. units
• B

  $\frac{1}{6}$ sq. units
• C

  $\frac{2}{3}$ sq. units
• D

  $\frac{6}{7}$ aq. units

Solution

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

Hence, the correct answer is option: 'A'.