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

Question

Aicharya · Accepted Answer

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

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

Solution