# integration by parts to evaluate the definite integral

• Feb 23rd 2010, 12:03 PM
dorrisstroy20
integration by parts to evaluate the definite integral
(9t^2)(ln(t))dt at points 1 to e
• Feb 23rd 2010, 12:07 PM
icemanfan
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 PM
dorrisstroy20
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 PM
e^(i*pi)
Quote:

Originally Posted by dorrisstroy20
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

$\displaystyle \int^e _1 (9t^2\, \ln(t)) \,dt = \int^e _1 \left[3t^3\, \ln(t) - \int 3t^3 \cdot \frac{1}{t}\right] = \left[3t^3\, \ln(t) - \frac{t^2}{3}\right]^e _1 = F(e) - F(1)$

Spoiler:
$\displaystyle \left(3e^3\, \ln(e) - \frac{e^2}{3}\right) - \left(3(1)^3\, \ln(1) - \frac{1^2}{3}\right)$
• Feb 23rd 2010, 12:24 PM
icemanfan
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 PM
dorrisstroy20
Thank you so much