(9t^2)(ln(t))dt at points 1 to e

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- Feb 23rd 2010, 12:03 PMdorrisstroy20integration by parts to evaluate the definite integral
(9t^2)(ln(t))dt at points 1 to e

- Feb 23rd 2010, 12:07 PMicemanfan
Choose $\displaystyle u = \ln t$ and $\displaystyle dv = 9t^2 \cdot dt$.

Then $\displaystyle du = \frac{1}{t} \cdot dt$ and $\displaystyle v = 3t^3$.

Thus, calculate:

$\displaystyle \int u \cdot dv = uv - \int v \cdot du$ - Feb 23rd 2010, 12:14 PMdorrisstroy20
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

- Feb 23rd 2010, 12:23 PMe^(i*pi)
- Feb 23rd 2010, 12:24 PMicemanfan
So then, you calculated

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

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

Hence

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

by the Fundamental Theorem of Calculus. Just simplify that expression. - Feb 23rd 2010, 12:33 PMdorrisstroy20
Thank you so much