# Substitution then integration by parts

• February 1st 2009, 06:12 PM
mollymcf2009
Substitution then integration by parts
Hey y'all,

The instructions for this problem are:

First make a substitution then evaluate the integral using integration by parts.

Here is what I did, can someone let me know if I'm on the right track?

$\int 2t^{9}e^{-t^5} dt$

$2 \int t^{9}e^{-t^5} dt$

$u = -t^5$
$du = -5t^4 dt$

$-\frac{1}{5} du = t^4 dt$

$-\frac{1}{5} \cdot 2 \int -u \cdot e^u du$

* It that what I'm supposed to do for the substitution part?
* If this is correct, do I plug my original numbers back in and then do my integration by parts the usual way?

Thanks!!
• February 1st 2009, 06:46 PM
Soroban
Hello, mollymcf2009!

Quote:

First make a substitution.
then evaluate the integral using integration by parts.

. . $\int 2t^9e^{-t^5}\,dt$

Here is what I did, can someone let me know if I'm on the right track?

$\int 2t^9e^{-t^5}\,dt \;=\;2 \int t^9e^{-t^5}\,dt$

. . $w \,=\, \text{-}t^5 \quad\Rightarrow\quad dw \,=\,\text{-}5t^4 dt \quad\Rightarrow\quad t^4 dt\,=\,\text{-}\frac{dw}{5}$

$-\frac{1}{5} \cdot 2 \int -w \cdot e^w dw$ . . . . correct, but tidy up!

You have: . $\frac{2}{5}\int w\!\cdot\!e^w\,dw$

By parts: . $\begin{array}{ccccccc}u &=& w & & dv &=& e^w\,dw \\ du &=& dw & & v &=& e^w \end{array}$

Then we have: . $\frac{2}{5}\left[w\!\cdot\!e^w - \int e^w\,dw\right]$ . . . etc.