calculus 2

find the definite integral for 3x+pi for the fraction interval pi over 3 and pi over 6

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asked by bern
  1. 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

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    posted by Reiny
  2. How to find the definite integral for (3x+pi)cos3xdx for the fraction interval pi over 6 and pi over 3

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

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    posted by Reiny

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