1. **State the problem:** Find the area under the curve $y = 3x^2 - 4x$ bounded by the vertical lines $x = -1$, $x = 2$, and the $x$-axis. 2. **Understand the problem:** The area under the curve and above the $x$-axis between $x = -1$ and $x = 2$ can be found by integrating the function and considering where the curve is above or below the $x$-axis. 3. **Find the points where the curve intersects the $x$-axis:** Solve $3x^2 - 4x = 0$. $$3x^2 - 4x = x(3x - 4) = 0$$ So, $x = 0$ or $x = \frac{4}{3}$. 4. **Split the integral at these points:** The interval $[-1, 2]$ is split into three parts: $[-1, 0]$, $[0, \frac{4}{3}]$, and $[\frac{4}{3}, 2]$. 5. **Determine the sign of the function on each interval:** - For $x$ in $[-1, 0]$, pick $x = -0.5$: $y = 3(-0.5)^2 - 4(-0.5) = 0.75 + 2 = 2.75 > 0$. - For $x$ in $[0, \frac{4}{3}]$, pick $x = 1$: $y = 3(1)^2 - 4(1) = 3 - 4 = -1 < 0$. - For $x$ in $[\frac{4}{3}, 2]$, pick $x = 1.5$: $y = 3(1.5)^2 - 4(1.5) = 6.75 - 6 = 0.75 > 0$. 6. **Set up the integral for the total area:** Since the function is negative on $[0, \frac{4}{3}]$, take the absolute value of the integral there. $$\text{Area} = \int_{-1}^0 (3x^2 - 4x) \, dx - \int_0^{\frac{4}{3}} (3x^2 - 4x) \, dx + \int_{\frac{4}{3}}^2 (3x^2 - 4x) \, dx$$ 7. **Compute each integral:** $$\int (3x^2 - 4x) \, dx = x^3 - 2x^2 + C$$ - From $-1$ to $0$: $$[x^3 - 2x^2]_{-1}^0 = (0 - 0) - ((-1)^3 - 2(-1)^2) = 0 - (-1 - 2) = 3$$ - From $0$ to $\frac{4}{3}$: $$\left[x^3 - 2x^2\right]_0^{\frac{4}{3}} = \left(\left(\frac{4}{3}\right)^3 - 2\left(\frac{4}{3}\right)^2\right) - 0 = \frac{64}{27} - 2 \times \frac{16}{9} = \frac{64}{27} - \frac{32}{9} = \frac{64}{27} - \frac{96}{27} = -\frac{32}{27}$$ - From $\frac{4}{3}$ to $2$: $$[x^3 - 2x^2]_{\frac{4}{3}}^2 = (8 - 8) - \left(\frac{64}{27} - \frac{32}{9}\right) = 0 - \left(\frac{64}{27} - \frac{96}{27}\right) = \frac{32}{27}$$ 8. **Calculate total area:** $$3 - \left(-\frac{32}{27}\right) + \frac{32}{27} = 3 + \frac{32}{27} + \frac{32}{27} = 3 + \frac{64}{27} = \frac{81}{27} + \frac{64}{27} = \frac{145}{27} \approx 5.37$$ **Final answer:** The area under the curve between $x = -1$ and $x = 2$ and above the $x$-axis is $$\boxed{\frac{145}{27}}$$.