Question

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

Answer 1

First you integrate the function:

`intx^3+2x^2-8x-1=1/4x^4+2/3x^3-4x^2-x`

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

`1/4(4)^4+2/3(4)^3-4(4)^2-4= 1/4(256)+2/3(64)-4(16)-4`

Solving that out yields:

`64+128/3-64-4= 116/3 (or 38.66666)`

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 `116/3 - 0`, which of course is just `116/3`, and that's your answer.