1. **State the problem:** We have the function $f(x) = (6 - 3x)(4 + x)$ and a shaded region $R$ bounded by the x-axis, y-axis, and the graph of $f$. We need to find: (a) An integral expression for the area of $R$. (b) The numerical value of the area of $R$. (c) The x-coordinate $a$ of point $C(a,0)$ such that the triangle $ABC$ with vertices $A(0,0)$, $B(3,10)$, and $C(a,0)$ has the same area as region $R$. 2. **Find the x-intercepts of $f(x)$ to determine the bounds of $R$: ** $$f(x) = (6 - 3x)(4 + x) = 24 + 6x - 12x - 3x^2 = 24 - 6x - 3x^2$$ Set $f(x) = 0$: $$24 - 6x - 3x^2 = 0 \implies -3x^2 - 6x + 24 = 0 \implies x^2 + 2x - 8 = 0$$ Solve quadratic: $$x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2}$$ So, $$x = 2 \quad \text{or} \quad x = -4$$ Since the region $R$ is bounded by the y-axis ($x=0$) and the curve, and above the x-axis, the relevant interval is $x \in [0,2]$. 3. **(a) Write down the integral for the area of $R$: ** The area under the curve from $x=0$ to $x=2$ above the x-axis is: $$\text{Area} = \int_0^2 f(x) \, dx = \int_0^2 (6 - 3x)(4 + x) \, dx$$ 4. **(b) Calculate the area of $R$: ** Expand the integrand: $$f(x) = (6 - 3x)(4 + x) = 24 + 6x - 12x - 3x^2 = 24 - 6x - 3x^2$$ Integrate term-by-term: $$\int_0^2 (24 - 6x - 3x^2) \, dx = \left[24x - 3x^2 - x^3\right]_0^2$$ Evaluate at $x=2$: $$24(2) - 3(2)^2 - (2)^3 = 48 - 12 - 8 = 28$$ Evaluate at $x=0$: $$0$$ So, area $= 28$. 5. **(c) Find $a$ such that the triangle $ABC$ has area equal to 28: ** Vertices: $A(0,0)$, $B(3,10)$, $C(a,0)$ Area of triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Base is along x-axis between $A$ and $C$, length $|a - 0| = a$ (since $a > 0$). Height is the y-coordinate of $B$, which is 10. Set area equal to 28: $$\frac{1}{2} \times a \times 10 = 28 \implies 5a = 28 \implies a = \frac{28}{5} = 5.6$$ **Final answers:** (a) $\displaystyle \int_0^2 (6 - 3x)(4 + x) \, dx$ (b) Area $= 28$ (c) $a = 5.6$