Mathematics - Area Under Curves Question with Solution | TestHub

Given,

$x-2y+4\ge 0, x+2y^2\ge 0, x+4y^2\le 8, y\ge 0$

Now, finding the point of intersection of $x-2y+4=0 \& x+2y^2=0$ we get,

$\left(x,y\right)\equiv \left(-2,1\right)$

And point of intersection of $x-2y+4=0 \& x+4y^2=8$ will be, $\left(x,y\right)\equiv \left(-1,\frac{3}{2}\right)$

Now, plotting the diagramw we get,

Now, from the above diagram, the required area will be,

$A={\int }_{-2}^{-1}\left(\frac{x+4}{2}-\sqrt{\frac{-x}{2}}\right)+{\int }_{-1}^0\left(\frac{\sqrt{8-x}}{2}-\sqrt{\frac{-x}{2}}\right)dx+{\int }_0^8\frac{\sqrt{8-x}}{2}dx$

$\Rightarrow A=\frac{1}{2}{\left[\frac{{\left(x+4\right)}^2}{2}\right]}_{-2}^{-1}-\frac{1}{\sqrt{2}}{\left[{\left(-x\right)}^{\frac{3}{2}}\left(\frac{-2}{3}\right)\right]}_{-2}^{-1}+\frac{1}{2}{\left[{\left(8-x\right)}^{\frac{3}{2}}\left(\frac{-2}{3}\right)\right]}_{-1}^0-\frac{1}{\sqrt{2}}{\left[{\left(-x\right)}^{\frac{3}{2}}\left(\frac{-2}{3}\right)\right]}_{-1}^0+\frac{1}{2}{\left[{\left(8-x\right)}^{\frac{3}{2}}\left(\frac{-2}{3}\right)\right]}_0^8$

$\Rightarrow A=9+\frac{5}{4}-\frac{4}{3}$

$\Rightarrow A=\frac{107}{12}$

$\Rightarrow m+n=119$