Integration by Parts problem

**Kimmy2** · November 9th 2009, 04:27 PM

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

Help is very much appreciated. Thank you!

**Prove It** · November 9th 2009, 04:32 PM

Originally Posted by Kimmy2

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

Help is very much appreciated. Thank you!

Recall that

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

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

Thus

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

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

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

Can you now find $\int_1^e{7t^2\ln{t}\,dt}$ ?

**BERRY** · November 9th 2009, 04:36 PM

For the second problem, set $u = t$ and $dv = e^{-t}dt$

**Kimmy2** · November 9th 2009, 05:19 PM

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?

**BERRY** · November 9th 2009, 05:54 PM

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
$<br /> \int{-e^{-t}dt} = e^{-t}$

**Kimmy2** · November 9th 2009, 07:04 PM

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?

**BERRY** · November 9th 2009, 07:32 PM

Yes that is correct. Just plug in the bounds now.

**Kimmy2** · November 9th 2009, 07:38 PM

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!!

**Prove It** · November 9th 2009, 07:51 PM

Originally Posted by Kimmy2

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

Help is very much appreciated. Thank you!

Q.1

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

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

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

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

Q.2

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

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

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

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