1. **Problem statement:** Find the area of the shaded region bounded by the curve $$y = x^3 - 39x - 70$$ and the x-axis between $$x = -5$$ and $$x = 7$$. The curve crosses the x-axis at $$x = -5$$, $$x = -2$$, and $$x = 7$$. 2. **Formula and approach:** The area between a curve and the x-axis from $$a$$ to $$b$$ is given by $$\int_a^b |f(x)| \, dx$$. Since the curve crosses the x-axis, we split the integral at the roots and take the absolute value of the function in each interval. 3. **Set up the integral:** - From $$x = -5$$ to $$x = -2$$, the curve is above the x-axis, so area is $$\int_{-5}^{-2} (x^3 - 39x - 70) \, dx$$. - From $$x = -2$$ to $$x = 7$$, the curve is below the x-axis, so area is $$\int_{-2}^{7} -(x^3 - 39x - 70) \, dx$$. 4. **Calculate the first integral:** $$\int (x^3 - 39x - 70) \, dx = \frac{x^4}{4} - \frac{39x^2}{2} - 70x + C$$ Evaluate from $$-5$$ to $$-2$$: $$\left[ \frac{x^4}{4} - \frac{39x^2}{2} - 70x \right]_{-5}^{-2} = \left( \frac{(-2)^4}{4} - \frac{39(-2)^2}{2} - 70(-2) \right) - \left( \frac{(-5)^4}{4} - \frac{39(-5)^2}{2} - 70(-5) \right)$$ Calculate each term: $$\frac{16}{4} - \frac{39 \times 4}{2} + 140 = 4 - 78 + 140 = 66$$ $$\frac{625}{4} - \frac{39 \times 25}{2} + 350 = 156.25 - 487.5 + 350 = 18.75$$ Subtract: $$66 - 18.75 = 47.25$$ 5. **Calculate the second integral:** $$\int -(x^3 - 39x - 70) \, dx = \int (-x^3 + 39x + 70) \, dx = -\frac{x^4}{4} + \frac{39x^2}{2} + 70x + C$$ Evaluate from $$-2$$ to $$7$$: $$\left[-\frac{x^4}{4} + \frac{39x^2}{2} + 70x \right]_{-2}^{7} = \left(-\frac{7^4}{4} + \frac{39 \times 7^2}{2} + 70 \times 7 \right) - \left(-\frac{(-2)^4}{4} + \frac{39 \times (-2)^2}{2} + 70 \times (-2) \right)$$ Calculate each term: $$-\frac{2401}{4} + \frac{39 \times 49}{2} + 490 = -600.25 + 955.5 + 490 = 845.25$$ $$-\frac{16}{4} + \frac{39 \times 4}{2} - 140 = -4 + 78 - 140 = -66$$ Subtract: $$845.25 - (-66) = 845.25 + 66 = 911.25$$ 6. **Total area:** $$47.25 + 911.25 = 958.5$$ **Final answer:** The area of the shaded region is $$\boxed{958.5}$$ square units.