Posted by Stacy on Friday, December 2, 2011 at 1:10am.
definite Integrals (using fundamental Theorem)
Evaluate
from 1 to 2(x^2  4x)dx

Calculus  drwls, Friday, December 2, 2011 at 1:38am
[x^3/3  2 x^2] @ 2
MINUS
[x^3/3  2 x^2] @ 1
= [8/3  8]  [1/3 2]
= 16/3 + 7/3 = 3

Calculus  Stacy, Friday, December 2, 2011 at 2:13am
Thank you so much. I had the integral part, but was off on the rest. :)
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