Calculus problem #2204

Since $y=x^3 + 6x^2 + 9x=x(x+3)^2$ , the curve intersects the x-axis at $x=-3$ and $x=0$ . Thus, $y \leq 0$ in the domain $-3 \leq x \leq 0$ . So, we have

$\begin{aligned} S &= - \int_{-3}^{0} y\ dx \\ &= \int_{0}^{-3} (x^3+6x^2+9x)\ dx \\ &= \left[ \frac{1}{4}x^4 + 2x^3+\frac{9}{2}x^2 \right]_{0}^{-3} \\ &= \frac{(-3)^4}{4} + 2\cdot (-3)^3 + \frac{9}{2}\cdot (-3)^2 \\ &= \frac{27}{4}. \\ \end{aligned}$