calculus 2
posted by bern .
find the definite integral for 3x+pi for the fraction interval pi over 3 and pi over 6

interpretation:
find ∫(3x + π) dx from π/3 to π/6
= [(3/2)x^2 + πx] from π/3 to π/6
= (3/2)(π^2/36) + π(π/6)  ((3/2)(π^2/9) + π(π/3)
= π^2 /24 + π^2/6  π^2 /6  π^2/3
= 7π/24 
How to find the definite integral for (3x+pi)cos3xdx for the fraction interval pi over 6 and pi over 3

http://www.wolframalpha.com/input/?i=%E2%88%AB(3x%2Bpi)(cos3x)+dx
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