Let for x ∈ R , f x = x + x 2 and g x = x , x 2 , x < 0 x ≥ 0 . Then area bounded by the…

We have,

$f\left(x\right)=\frac{x+\left|x\right|}{2}=\left\{\begin{array}{l}\begin{array}{ll}x;& x\ge 0\end{array}\\ 0; x<0\end{array}\right.$

$g\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}{x}^2;& x\ge 0\end{array}\\ x; x<0\end{array}\right.$

So,

$fog\left(x\right)=f\left\{g\left(x\right)\right\}=\left[\begin{array}{cc}g\left(x\right);& g\left(x\right)\ge 0\\ 0;& g\left(x\right)<0\end{array}\right.$

$\Rightarrow fog\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}{x}^2;& x\ge 0\end{array}\\ 0; x<0\end{array}\right.$

And,

$2y-x=15$

Solving $2y-x=15$ and $y=x^2$, we get

$2x^2-x-15=0$

$\Rightarrow x=\frac{1\pm 11}{4}$

$\Rightarrow x=3, -\frac{5}{2}$

Required area

$A={\int }_0^3\left(\frac{x+15}{2}-x^2\right)\mathrm{dx}+\frac{1}{2}\times \frac{15}{2}\times 15$

$\Rightarrow A={\left[\frac{x^2}{4}+\left(\frac{15}{2}\right)x-\frac{x^3}{3}\right]}_0^3 + \frac{225}{4}$

$\Rightarrow A=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}$

$\Rightarrow A=\frac{99-36+225}{4}$

$\Rightarrow A=72$