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Math Help - integration by parts to evaluate the definite integral

  1. #1
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    integration by parts to evaluate the definite integral

    (9t^2)(ln(t))dt at points 1 to e
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  2. #2
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    Choose u = \ln t and dv = 9t^2 \cdot dt.
    Then du = \frac{1}{t} \cdot dt and v = 3t^3.

    Thus, calculate:

    \int u \cdot dv = uv - \int v \cdot du
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  3. #3
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    first thanks for your response and second I got that far but I am having trouble evaluating it at the points from 1 to e
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  4. #4
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    Quote Originally Posted by dorrisstroy20 View Post
    first thanks for your response and second I got that far but I am having trouble evaluating it at the points from 1 to e
    \int^e _1 (9t^2\, \ln(t)) \,dt = \int^e _1 \left[3t^3\, \ln(t) - \int 3t^3 \cdot \frac{1}{t}\right] = \left[3t^3\, \ln(t) - \frac{t^2}{3}\right]^e _1 = F(e) - F(1)

    Spoiler:
    \left(3e^3\, \ln(e) - \frac{e^2}{3}\right) - \left(3(1)^3\, \ln(1) - \frac{1^2}{3}\right)
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  5. #5
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    So then, you calculated

    \int 9t^2 \ln t \cdot dt = 3t^3 \ln t - \int 3t^2 \cdot dt

    \int 9t^2 \ln t \cdot dt = 3t^3 \ln t - t^3 + C.

    Hence

    \int _1 ^e 9t^2 \ln t \cdot dt = 3e^3 \ln e - e^3 - (3 \cdot 1^3 \ln 1 - 1^3),

    by the Fundamental Theorem of Calculus. Just simplify that expression.
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  6. #6
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    Thank you so much
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