# How do you evaluate the integral int_0^4x^3+2x^2-8x-1?

Oct 14, 2014

First you integrate the function:

$\int {x}^{3} + 2 {x}^{2} - 8 x - 1 = \frac{1}{4} {x}^{4} + \frac{2}{3} {x}^{3} - 4 {x}^{2} - x$

Then you substitute in your values for the upper and lower bounds. Start with 4:

$\frac{1}{4} {\left(4\right)}^{4} + \frac{2}{3} {\left(4\right)}^{3} - 4 {\left(4\right)}^{2} - 4 = \frac{1}{4} \left(256\right) + \frac{2}{3} \left(64\right) - 4 \left(16\right) - 4$

Solving that out yields:

$64 + \frac{128}{3} - 64 - 4 = \frac{116}{3} \left(\mathmr{and} 38.66666\right)$

Next you would substitute in 0, but looking at the equation, you can see that subbing 0 in will just yield zero. So last you do $\frac{116}{3} - 0$, which of course is just $\frac{116}{3}$, and that's your answer.