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

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1 Answer
Johnny L.
Oct 14, 2014

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.

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