### Home > APCALC > Chapter 4 > Lesson 4.3.2 > Problem4-120

4-120.

Calculate the area of the region in the second quadrant under the curve $y = x^{3} + 2x^{2} – 3x$.

To set up the integral, first determine the bounds in which this function exists in the 2nd quadrant.

Determine the roots:
$0 = x^{3} + 2x^{2} − 3x$
$0 = x\left(x^{2} + 2x − 3\right) = x\left(x + 3\right)\left(x − 1\right)$
Roots: $x = 0, x = −3$, and $x = 1$
2nd quadrant domain: $[−3, 0]$

Integrate.

$\int_{-3}^{0} (x^3+2x^2-3x) dx=\\ = \tfrac{1}{4}x^4+\frac{2}{3}x^3-\frac{3}{2}x^2\Big|_{-3}^0=11.25$