Let for x ∈ R , f(x) = (x + |x|)/(2) and g(x) = \ array ll x, x ^ 2...

We have,

$f$\left(x\right) = \dfrac{x + \left|x\right|}{2} = \left\{$\begin{array}{ll}x; x \ge 0\end{array}$0; x\right)</p><p>g\left(x\right) = \left\{$$\begin{array}{ll}{x}^{2}; x \ge 0\end{array}$$ x; x\right)

So,

fog\left(x\right) = f\left\{g\left(x\right)\right\} = \left[\begin{cases}g\left(x\right); & g\left(x\right) \ge 0 \\ 0; & g\left(x\right) < 0\end{cases}\right)</p><p>⇒ fog\left(x\right) = \left\{$$\begin{array}{ll}{x}^{2}; x \ge 0\end{array}$$ 0; x\right)

And,

$2y - x = 15$

Solving $2y - x = 15$ and $y = {x}^{2}$, we get

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

$⇒ x = \dfrac{1 \pm 11}{4}$

$⇒ x = 3, - \dfrac{5}{2}$

Required area

$A = {∫}_{0}^{3}\left(\dfrac{x + 15}{2} - {x}^{2}\right)dx + \dfrac{1}{2} \times \dfrac{15}{2} \times 15$

$⇒ A = {\left[\dfrac{{x}^{2}}{4} + \left(\dfrac{15}{2}\right)x - \dfrac{{x}^{3}}{3}\right]}_{0}^{3} + \dfrac{225}{4}$

$⇒ A = \dfrac{9}{4} + \dfrac{45}{2} - 9 + \dfrac{225}{4}$

$⇒ A = \dfrac{99 - 36 + 225}{4}$

$⇒ A = 72$