Mathematics - Area Under Curves Question with Solution | TestHub

Point of intersection of $y^2=x$ and $x+y=2$ is $$(2-x{)}^2=x\Rightarrow x^2-4x-x+4=0$$ $$x^2-5x+4=0\Rightarrow (x-4)(x-1)=0\Rightarrow x=1,4$$ 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 $$\begin{array}{l}={\int }_0^1\sqrt{x}dx+{\int }_1^2(2-x)dx\\ =\frac{2}{3}[x\sqrt{x}{]}_0^1+2[x{]}_1^2-\frac{1}{2}{\left[x^2\right]}_1^2\\ =\frac{2}{3}+2-\left(\frac{1}{2}\times 3\right)=\frac{2}{3}+2-\frac{3}{2}=\frac{4+12-9}{6}=\frac{7}{6}\end{array}$$