1. **Problem statement:** Calculate the area of the region bounded by the curve of the inverse function $f^{-1}$, the line $y=x$, and the vertical lines $x=\alpha$ and $x=0$ by deducing it from the curve of $f$. 2. **Recall the property of areas between inverse functions:** If $f$ is invertible and continuous on $[\alpha, +\infty[$, the area between $f^{-1}$ and the line $y=x$ from $x=\alpha$ to $x=0$ can be found using the integral of $f$: $$\text{Area} = \int_{\alpha}^0 f(x)\,dx - \int_{f(\alpha)}^{f(0)} y\,dy$$ But since $f(0)$ and $f(\alpha)$ are the $y$-values corresponding to $x=0$ and $x=\alpha$, and the region is bounded by $x=\alpha$ and $x=0$, the area between $f^{-1}$ and $y=x$ is: $$\text{Area} = \int_{\alpha}^0 f(x)\,dx - \frac{1}{2}[f(0)^2 - f(\alpha)^2]$$ 3. **Calculate the needed values:** - $f(0) = (-0 - 2)e^{0} + 0 = -2 + 0 = -2$ - $f(\alpha)$ with $\alpha = -1.28$: $$f(-1.28) = (-(-1.28) - 2)e^{1.28} + (-1.28) = (1.28 - 2)e^{1.28} - 1.28 = (-0.72)e^{1.28} - 1.28$$ Calculate $e^{1.28} \approx 3.6$ (approximate): $$f(-1.28) \approx -0.72 \times 3.6 - 1.28 = -2.592 - 1.28 = -3.872$$ 4. **Calculate the integral $\int_{\alpha}^0 f(x) dx$:** Recall $f(x) = (-x - 2)e^{-x} + x$. Rewrite: $$f(x) = (-x - 2)e^{-x} + x = -(x+2)e^{-x} + x$$ Integral: $$\int_{\alpha}^0 f(x) dx = \int_{\alpha}^0 [-(x+2)e^{-x} + x] dx = -\int_{\alpha}^0 (x+2)e^{-x} dx + \int_{\alpha}^0 x dx$$ Calculate each separately. 5. **Integral $\int (x+2)e^{-x} dx$:** Use integration by parts: Let $I = \int (x+2)e^{-x} dx = \int x e^{-x} dx + 2 \int e^{-x} dx$ - $\int e^{-x} dx = -e^{-x} + C$ - For $\int x e^{-x} dx$, use integration by parts: Set $u = x$, $dv = e^{-x} dx$, then $du = dx$, $v = -e^{-x}$ $$\int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$$ So, $$I = -e^{-x}(x+1) - 2 e^{-x} + C = -e^{-x}(x+3) + C$$ 6. **Evaluate definite integral:** $$\int_{\alpha}^0 (x+2)e^{-x} dx = [-e^{-x}(x+3)]_{\alpha}^0 = [-e^{0}(0+3)] - [-e^{-\alpha}(\alpha + 3)] = -3 + e^{-\alpha}(\alpha + 3)$$ 7. **Calculate $\int_{\alpha}^0 x dx$:** $$\int_{\alpha}^0 x dx = \left[ \frac{x^2}{2} \right]_{\alpha}^0 = 0 - \frac{\alpha^2}{2} = -\frac{\alpha^2}{2}$$ 8. **Combine results:** $$\int_{\alpha}^0 f(x) dx = -\left(-3 + e^{-\alpha}(\alpha + 3)\right) - \frac{\alpha^2}{2} = 3 - e^{-\alpha}(\alpha + 3) - \frac{\alpha^2}{2}$$ 9. **Calculate the area:** $$\text{Area} = \int_{\alpha}^0 f(x) dx - \frac{1}{2}[f(0)^2 - f(\alpha)^2] = 3 - e^{-\alpha}(\alpha + 3) - \frac{\alpha^2}{2} - \frac{1}{2}[(-2)^2 - (-3.872)^2]$$ Calculate the square terms: $$(-2)^2 = 4$$ $$(-3.872)^2 \approx 15.0$$ So, $$\frac{1}{2}[4 - 15.0] = \frac{1}{2}(-11) = -5.5$$ 10. **Final area:** $$\text{Area} = 3 - e^{-\alpha}(\alpha + 3) - \frac{\alpha^2}{2} + 5.5 = 8.5 - e^{-\alpha}(\alpha + 3) - \frac{\alpha^2}{2}$$ Substitute $\alpha = -1.28$ and $e^{-\alpha} = e^{1.28} \approx 3.6$: $$e^{-\alpha}(\alpha + 3) = 3.6 \times (-1.28 + 3) = 3.6 \times 1.72 = 6.192$$ $$\frac{\alpha^2}{2} = \frac{(-1.28)^2}{2} = \frac{1.6384}{2} = 0.8192$$ Therefore, $$\text{Area} \approx 8.5 - 6.192 - 0.8192 = 8.5 - 7.0112 = 1.4888$$ **Answer:** The area of the region bounded by $f^{-1}$, the line $y=x$, and the vertical lines $x=\alpha$ and $x=0$ is approximately **1.49**.