# Integration by Parts problem

• Nov 9th 2009, 04:27 PM
Kimmy2
Integration by Parts problem
I've been struggling with integration by parts. Can anyone explain these problems to me?

http://hosted2.webwork.rochester.edu...361817ca91.png

http://hosted2.webwork.rochester.edu...e3b5345091.png

Help is very much appreciated. Thank you!
• Nov 9th 2009, 04:32 PM
Prove It
Quote:

Originally Posted by Kimmy2
I've been struggling with integration by parts. Can anyone explain these problems to me?

http://hosted2.webwork.rochester.edu...361817ca91.png

http://hosted2.webwork.rochester.edu...e3b5345091.png

Help is very much appreciated. Thank you!

Recall that

$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

For Q.1 let $\displaystyle u = \ln{t}$ so that $\displaystyle du = \frac{1}{t}$, and let $\displaystyle dv = 7t^2$ so that $\displaystyle v = \frac{7}{3}t^3$.

Thus

$\displaystyle \int{7t^2 \ln{t}\,dt} = \frac{7}{3}t^3\ln{t} - \int{\frac{7}{3}t^3\,\frac{1}{t}\,dt}$

$\displaystyle = \frac{7}{3}t^3\ln{t} - \frac{7}{3}\int{t^2\,dt}$

$\displaystyle = \frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3 + C$.

Can you now find $\displaystyle \int_1^e{7t^2\ln{t}\,dt}$?
• Nov 9th 2009, 04:36 PM
BERRY
For the second problem, set $\displaystyle u = t$ and $\displaystyle dv = e^{-t}dt$
• Nov 9th 2009, 05:19 PM
Kimmy2
So, I have:

u=t
du=dx

dv=e^-t
v= -e^(-t)

And then:

-te^(-t) - the integral from 6 to 0 of -e^(-t)dx

And then if I take the integral of -e^-tdx, I get -ln(e^-t)?

Is this okay, so far?
• Nov 9th 2009, 05:54 PM
BERRY
Quote:

Originally Posted by Kimmy2
So, I have:

u=t
du=dx

dv=e^-t
v= -e^(-t)

And then:

-te^(-t) - the integral from 6 to 0 of -e^(-t)dx

And then if I take the integral of -e^-tdx, I get -ln(e^-t)?

Is this okay, so far?

A few issues.

u = t so du = dt, not dx
Also
$\displaystyle \int{-e^{-t}dt} = e^{-t}$
• Nov 9th 2009, 07:04 PM
Kimmy2
So, after I plug in all the u and v values, I'm getting:

t(-e^-t) - the integral of -e^(-t)

Once I integrate that last part, I'm getting:

t(-e^-t) - e^(-t)

Is this right?
• Nov 9th 2009, 07:32 PM
BERRY
Yes that is correct. Just plug in the bounds now.
• Nov 9th 2009, 07:38 PM
Kimmy2
I have plugged in the numbers multiple times, and my homework keeps telling me that I'm wrong. :(

Some of the answers I got were:
-0.1983001 and 1.01239376

EDIT!
I got it, thanks so much for your help!!
• Nov 9th 2009, 07:51 PM
Prove It
Quote:

Originally Posted by Kimmy2
I've been struggling with integration by parts. Can anyone explain these problems to me?

http://hosted2.webwork.rochester.edu...361817ca91.png

http://hosted2.webwork.rochester.edu...e3b5345091.png

Help is very much appreciated. Thank you!

Q.1

$\displaystyle \int_1^e{7t^2\ln{t}\,dt} = \left[\frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3\right]_1^e$

$\displaystyle = \left[\frac{7}{3}e^3\ln{e} - \frac{7}{9}e^3\right] - \left[\frac{7}{3}(1)^3 \ln{1} - \frac{7}{9}(1)^3\right]$

$\displaystyle = \left[\frac{7}{3}e^3 - \frac{7}{9}e^3\right] - \left[- \frac{7}{9}\right]$

$\displaystyle = \frac{14}{9}e^3 + \frac{7}{9}$.

Q.2

$\displaystyle \int_0^6{t e^{-t}\,dt} = \left[-te^{-t} - e^{-t}\right]_0^6$

$\displaystyle = \left[-6e^{-6} - e^{-6}\right] - \left[-0e^{-0} - e^{-0}\right]$

$\displaystyle = -7e^{-6} - 1$.