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**R3ap3r** Find the exact area of the region between the curve $\displaystyle y = 4x^3 - 2x^2 - 12x$ and the x=axis, over the interval [-1, 2]

My Work:

$\displaystyle \int_{-1}^2\;(4x^3 - 2x^2 - 12x)\;dx = x^4 - \frac{2}{3}\; x^3 - 6x^2$

$\displaystyle Let\; x = 2\;\;\;\;\; 2^4 - \frac{2}{3}\; (2)^3 - 6(2)^2$

$\displaystyle Let\; x = -1\;\;\;\;\; (-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2$

$\displaystyle \left[2^4 - \frac{2}{3}\;(2)^3 - 6(2)^2\right] - \left[(-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2\right]$

$\displaystyle = 3.75\; units^2$

Hows my answer look here?