Integration By Parts: uv-Substitution - 9.2

[karush]

$\tiny{9.2}$
\begin{align*} \displaystyle
I&=\int y^3e^{-9y} \, dx\\
\textit{uv substitution}\\
u&=y^3\therefore \frac{1}{3}du=y^2dx\\
dv&=e^{-9y}\, dx\therefore v=e^{-9y}\\
\end{align*}
will stop there this looks like tabular method better

 

[MarkFL]

I moved your post to a new thread...:D

Did you intend for the integrand to be a function of $y$ while the differential is $dx$?

 

[topsquark]

$\tiny{9.2}$
\begin{align*} \displaystyle
I&=\int y^3e^{-9y} \, dx\\
\textit{uv substitution}\\
u&=y^3\therefore \frac{1}{3}du=y^2dx\\
dv&=e^{-9y}\, dx\therefore v=e^{-9y}\\
\end{align*}
will stop there this looks like tabular method better
opps should have started a new thread

I'd start off with the substitution z = 9y. Then the original integral becomes:
\(\displaystyle \int y^3 ~ e^{-9y} ~ dy = \left ( \frac{1}{9} \right )^4 \int z^3 e^{-z}~dz\)

Now do your integration by parts.

-Dan

 

[karush]

not sure about this
\begin{array}{rcl}
u&&v\\
z^3 &+&e^{-z} \\
&↘&\\
3z^2&- &e^{-z}\\
&↘&\\
6z &+ &e^{-z} \\
&↘&\\
6&-&e^{-z}
\end{array}

 

[karush]

\begin{array}{rcl}
u&&v\\
z^3 &+&e^{-z} \\
&\searrow &\\
3z^2&- &e^{-z}\\
&\searrow\\
6z &+ &e^{-z} \\
&\searrow&\\
6&-&e^{-z}
\end{array}
$\textsf{substitute $z=9y$}\\$
$$\dfrac{\left(243y^3+81y^2+18y+2\right)\mathrm{e}^{-9y}}{2187}$$