1. **Problem:** Find the area of the region bounded by the graph of $f(x) = x^3 - 1$ and the x-axis on the interval $[-1, 2]$. 2. **Formula:** The area bounded by the curve and the x-axis on $[a,b]$ is given by the definite integral of the absolute value of the function: $$\text{Area} = \int_a^b |f(x)| \, dx$$ 3. **Step 1: Find where $f(x)$ crosses the x-axis** Solve $x^3 - 1 = 0$: $$x^3 = 1 \implies x = 1$$ This means the function changes sign at $x=1$. 4. **Step 2: Split the integral at $x=1$** Calculate: $$\int_{-1}^1 |x^3 - 1| \, dx + \int_1^2 |x^3 - 1| \, dx$$ 5. **Step 3: Determine the sign of $f(x)$ on each interval** - For $x \in [-1,1]$, $f(x) = x^3 - 1 \leq 0$ (since $x^3 \leq 1$ and subtracting 1 makes it negative or zero). - For $x \in [1,2]$, $f(x) = x^3 - 1 \geq 0$. 6. **Step 4: Write the integrals without absolute values accordingly** $$\int_{-1}^1 -(x^3 - 1) \, dx + \int_1^2 (x^3 - 1) \, dx = \int_{-1}^1 (1 - x^3) \, dx + \int_1^2 (x^3 - 1) \, dx$$ 7. **Step 5: Compute each integral** - First integral: $$\int_{-1}^1 (1 - x^3) \, dx = \left[ x - \frac{x^4}{4} \right]_{-1}^1 = \left(1 - \frac{1}{4}\right) - \left(-1 - \frac{1}{4}\right) = \frac{3}{4} + 1 + \frac{1}{4} = 2$$ - Second integral: $$\int_1^2 (x^3 - 1) \, dx = \left[ \frac{x^4}{4} - x \right]_1^2 = \left( \frac{16}{4} - 2 \right) - \left( \frac{1}{4} - 1 \right) = (4 - 2) - \left( \frac{1}{4} - 1 \right) = 2 - \left( -\frac{3}{4} \right) = 2 + \frac{3}{4} = \frac{11}{4}$$ 8. **Step 6: Add the two areas** $$\text{Area} = 2 + \frac{11}{4} = \frac{8}{4} + \frac{11}{4} = \frac{19}{4} = 4.75$$ **Final answer:** The area bounded by the graph of $f(x) = x^3 - 1$ and the x-axis on $[-1, 2]$ is $$\boxed{\frac{19}{4}}$$ or 4.75.