1. **State the problem:** Find the area of the region bounded by the graph of the function $f(x) = x^3 + 1$, the x-axis, and the vertical lines $x = -2$ and $x = 0$. 2. **Formula and rules:** The area bounded by a curve $y = f(x)$ and the x-axis between $x = a$ and $x = b$ is given by the definite integral $$\text{Area} = \int_a^b |f(x)| \, dx.$$ Since the function may be above or below the x-axis, we take the absolute value to ensure the area is positive. 3. **Analyze the function:** Evaluate $f(x)$ at the boundaries: - At $x = -2$: $f(-2) = (-2)^3 + 1 = -8 + 1 = -7$ (below x-axis) - At $x = 0$: $f(0) = 0 + 1 = 1$ (above x-axis) Find where $f(x) = 0$ between $-2$ and $0$: $$x^3 + 1 = 0 \implies x^3 = -1 \implies x = -1.$$ This is the point where the curve crosses the x-axis. 4. **Split the integral at $x = -1$:** - From $x = -2$ to $x = -1$, $f(x) < 0$, so area contribution is $$\int_{-2}^{-1} -f(x) \, dx = \int_{-2}^{-1} -(x^3 + 1) \, dx.$$ - From $x = -1$ to $x = 0$, $f(x) > 0$, so area contribution is $$\int_{-1}^0 f(x) \, dx = \int_{-1}^0 (x^3 + 1) \, dx.$$ 5. **Calculate each integral:** $$\int (x^3 + 1) \, dx = \frac{x^4}{4} + x + C.$$ - For $x$ in $[-2, -1]$: $$\int_{-2}^{-1} -(x^3 + 1) \, dx = -\left[ \frac{x^4}{4} + x \right]_{-2}^{-1} = -\left( \left(\frac{(-1)^4}{4} + (-1)\right) - \left(\frac{(-2)^4}{4} + (-2)\right) \right).$$ Calculate inside: $$\frac{1}{4} - 1 = -\frac{3}{4}, \quad \frac{16}{4} - 2 = 4 - 2 = 2.$$ So the integral is: $$-\left(-\frac{3}{4} - 2\right) = -\left(-\frac{11}{4}\right) = \frac{11}{4} = 2.75.$$ - For $x$ in $[-1, 0]$: $$\int_{-1}^0 (x^3 + 1) \, dx = \left[ \frac{x^4}{4} + x \right]_{-1}^0 = \left(0 + 0\right) - \left(\frac{1}{4} - 1\right) = 0 - \left(-\frac{3}{4}\right) = \frac{3}{4} = 0.75.$$ 6. **Add the two areas:** $$\text{Total area} = 2.75 + 0.75 = 3.5.$$ **Final answer:** The area of the region bounded by the curve, the x-axis, and the lines $x = -2$ and $x = 0$ is $\boxed{3.5}$ square units.