1. **State the problem:** We need to find the area of the region bounded by the curve $y = x^3 - 4x$ and the x-axis. 2. **Find the points where the curve intersects the x-axis:** Set $y = 0$. $$x^3 - 4x = 0$$ Factor out $x$: $$x(x^2 - 4) = 0$$ This gives: $$x = 0, \quad x^2 - 4 = 0 \Rightarrow x = \pm 2$$ 3. **Determine the intervals for integration:** The curve crosses the x-axis at $x = -2, 0, 2$. We will find the area between these points. 4. **Set up the integral for the area:** The area is the integral of the absolute value of the function between the roots: $$\text{Area} = \int_{-2}^0 |x^3 - 4x| \, dx + \int_0^2 |x^3 - 4x| \, dx$$ 5. **Determine the sign of $y$ on each interval:** - For $x$ in $[-2,0]$, pick $x = -1$: $(-1)^3 - 4(-1) = -1 + 4 = 3 > 0$, so $y > 0$. - For $x$ in $[0,2]$, pick $x = 1$: $1 - 4 = -3 < 0$, so $y < 0$. 6. **Rewrite the integral without absolute values:** $$\text{Area} = \int_{-2}^0 (x^3 - 4x) \, dx - \int_0^2 (x^3 - 4x) \, dx$$ 7. **Calculate each integral:** $$\int (x^3 - 4x) \, dx = \frac{x^4}{4} - 2x^2 + C$$ Evaluate from $-2$ to $0$: $$\left[ \frac{x^4}{4} - 2x^2 \right]_{-2}^0 = \left(0 - 0\right) - \left(\frac{16}{4} - 2 \times 4\right) = 0 - (4 - 8) = 4$$ Evaluate from $0$ to $2$: $$\left[ \frac{x^4}{4} - 2x^2 \right]_0^2 = \left(\frac{16}{4} - 2 \times 4\right) - (0 - 0) = (4 - 8) - 0 = -4$$ 8. **Calculate total area:** $$\text{Area} = 4 - (-4) = 8$$ **Final answer:** The area of the region bounded by the curve and the x-axis is $8$ square units.