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Step 1: Define the given functions $f(x)$ and $g(x)$ in piecewise form. The function $f(x)$ is given by $f(x) = \dfrac{x+|x|}{2}$. We can rewrite $f(x)$ based on the sign of $x$: If $x \geq 0$, then $|x|=x$, so $f(x) = \dfrac{x+x}{2} = \dfrac{2x}{2} = x$. If $x < 0$, then $|x|=-x$, so $f(x) = \dfrac{x-x}{2} = \dfrac{0}{2} = 0$. Thus, $f(x)$ can be written as: $$ f(x) = \begin{cases} x, & x \geq 0 \\ 0, & x < 0 \end{cases} $$ The function $g(x)$ is already given in piecewise form: $$ g(x) = \begin{cases} x, & x<0 \\ x^2, & x\geq0 \end{cases} $$ Step 2: Determine the composite function $(f \circ g)(x)$. The composite function $(f \circ g)(x) = f(g(x))$. We need to evaluate this based on the definition of $g(x)$. Case 1: For $x < 0$. In this case, $g(x) = x$. Since $x < 0$, we have $g(x) < 0$. Using the definition of $f(y)$ for $y < 0$, we get $f(y)=0$. So, $f(g(x)) = f(x) = 0$ for $x < 0$. Case 2: For $x \geq 0$. In this case, $g(x) = x^2$. Since $x \geq 0$, we have $x^2 \geq 0$, so $g(x) \geq 0$. Using the definition of $f(y)$ for $y \geq 0$, we get $f(y)=y$. So, $f(g(x)) = f(x^2) = x^2$ for $x \geq 0$. Combining these two cases, the function $y = (f \circ g)(x)$ is: $$ y = \begin{cases} 0, & x < 0 \\ x^2, & x \geq 0 \end{cases} $$ Step 3: Identify all bounding curves. The area is bounded by the curve $y=(f \circ g)(x)$, the line $y=0$ (the x-axis), and the line $2y-x=15$. Let's rewrite the line $2y-x=15$ in terms of $y$: $y = \dfrac{x+15}{2}$. Step 4: Find the intersection points of the bounding curves. We need to find where these curves intersect to determine the limits of integration. 1. Intersection of $y=0$ and $y=\dfrac{x+15}{2}$: $0 = \dfrac{x+15}{2} \implies x+15=0 \implies x=-15$. This gives the point $(-15, 0)$. This point satisfies $x < 0$, where $y=(f \circ g)(x)$ is $y=0$. 2. Intersection of $y=x^2$ (for $x \geq 0$) and $y=\dfrac{x+15}{2}$: $x^2 = \dfrac{x+15}{2}$ $2x^2 = x+15$ $2x^2 - x - 15 = 0$ Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$: $x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-15)}}{2(2)}$ $x = \dfrac{1 \pm \sqrt{1 + 120}}{4}$ $x = \dfrac{1 \pm \sqrt{121}}{4}$ $x = \dfrac{1 \pm 11}{4}$ Two possible values for $x$: $x_1 = \dfrac{1+11}{4} = \dfrac{12}{4} = 3$ and $x_2 = \dfrac{1-11}{4} = \dfrac{-10}{4} = -\dfrac{5}{2}$. Since we are considering $x \geq 0$ for $y=x^2$, we take $x=3$. For $x=3$, $y=x^2=3^2=9$. This gives the point $(3, 9)$. The region is bounded by $y=0$ (from $x=-15$ to $x=0$), $y=x^2$ (from $x=0$ to $x=3$), and the line $y=\dfrac{x+15}{2}$. The vertices of the enclosed region are $(-15,0)$, $(0,0)$, and $(3,9)$. Step 5: Set up the definite integral for the area. The area can be calculated by integrating with respect to $x$. We need to split the integral into two parts because the lower boundary curve $y=(f \circ g)(x)$ changes at $x=0$. For $x \in [-15, 0]$, the upper boundary is $y=\dfrac{x+15}{2}$ and the lower boundary is $y=0$. For $x \in [0, 3]$, the upper boundary is $y=\dfrac{x+15}{2}$ and the lower boundary is $y=x^2$. The total area $A$ is given by: $$ A = \int_{-15}^{0} \left( \dfrac{x+15}{2} - 0 \right) dx + \int_{0}^{3} \left( \dfrac{x+15}{2} - x^2 \right) dx $$ Step 6: Evaluate the definite integrals. Evaluate the first integral: $$ \int_{-15}^{0} \dfrac{x+15}{2} dx = \dfrac{1}{2} \left[ \dfrac{x^2}{2} + 15x \right]_{-15}^{0} $$ $$ = \dfrac{1}{2} \left[ \left( \dfrac{0^2}{2} + 15(0) \right) - \left( \dfrac{(-15)^2}{2} + 15(-15) \right) \right] $$ $$ = \dfrac{1}{2} \left[ 0 - \left( \dfrac{225}{2} - 225 \right) \right] = \dfrac{1}{2} \left[ - \left( -\dfrac{225}{2} \right) \right] = \dfrac{225}{4} $$ Evaluate the second integral: $$ \int_{0}^{3} \left( \dfrac{x+15}{2} - x^2 \right) dx = \left[ \dfrac{x^2}{4} + \dfrac{15x}{2} - \dfrac{x^3}{3} \right]_{0}^{3} $$ $$ = \left( \dfrac{3^2}{4} + \dfrac{15(3)}{2} - \dfrac{3^3}{3} \right) - \left( \dfrac{0^2}{4} + \dfrac{15(0)}{2} - \dfrac{0^3}{3} \right) $$ $$ = \left( \dfrac{9}{4} + \dfrac{45}{2} - \dfrac{27}{3} \right) - 0 $$ $$ = \left( \dfrac{9}{4} + \dfrac{90}{4} - \dfrac{36}{4} \right) = \dfrac{9+90-36}{4} = \dfrac{63}{4} $$ Add the results of the two integrals to find the total area: $$ A = \dfrac{225}{4} + \dfrac{63}{4} = \dfrac{225+63}{4} = \dfrac{288}{4} = 72 $$ Step 7: State the final answer. The area bounded by the given curves is $72$ square units. The final answer is $\boxed{72}$.