### Home > CALC > Chapter 4 > Lesson 4.3.2 > Problem4-109

4-109.

Find the area of the region in the second quadrant under the function $y = x^3 + 2x^2-3x$.

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

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

Integrate.

$\int_{-3}^0(x^3+2x^2-3x)dx=$