# Thread: Integration by Parts problem

1. ## Integration by Parts problem

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

Help is very much appreciated. Thank you!

2. 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 $\int_1^e{7t^2\ln{t}\,dt}$?

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

4. 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?

5. 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
$
\int{-e^{-t}dt} = e^{-t}$

6. 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?

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

8. 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!!

9. 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

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

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

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