Let for $$x \in R$$, $$f(x) = \dfrac{x+x}{2}$$ and $$g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \ge 0 \end{cases}$$. Then area bounded by the curve $$y = f \circ g(x)$$ and the lines $$y = 0, 2y - x = 15$$ is equal to ______.

For $$x < 0 \Rightarrow |x| = -x \Rightarrow f(x) = \frac{x - x}{2} = 0$$

For $$x \ge 0 \Rightarrow |x| = x \Rightarrow f(x) = \frac{x + x}{2} = x$$

$$x < 0$$:

$$\Rightarrow g(x) = x$$

Since $$x < 0$$, $$g(x) < 0$$. Therefore, $$f(g(x)) = 0$$.

$$\Rightarrow y = 0$$

$$x \ge 0$$:

$$\Rightarrow g(x) = x^2$$

Since $$x^2 \ge 0$$, $$g(x) \ge 0$$. Therefore, $$f(g(x)) = g(x) = x^2$$.

$$\Rightarrow y = x^2$$

$$y = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$$

The curve intersects the line at $$(3, 9)$$ and the line intersects the x axis at $$(-15,0)$$

$$\text{Area} = \int_{0}^{9} (x_{\text{right}} - x_{\text{left}}) \, dy$$

$$\Rightarrow \text{Area} = \int_{0}^{9} \left( \sqrt{y} - (2y - 15) \right) \, dy$$

$$\Rightarrow \text{Area} = \int_{0}^{9} \left( y^{1/2} - 2y + 15 \right) \, dy$$

$$\Rightarrow \text{Area} = \left[ \frac{2}{3}y^{3/2} - y^2 + 15y \right]_{0}^{9}$$

$$\Rightarrow \text{Area} = \left( \frac{2}{3}(9)^{3/2} - (9)^2 + 15(9) \right) - 0$$

$$\Rightarrow \text{Area} = 18 - 81 + 135 = 72$$