integral of x^3 * e^x dx

[math]

integral of x^3 * e^x dx

u= x^3 v' = e^x
u' = 3x^2 v=e^x

x^3 * e^x - integral of e^x * 3x^2

 

[soroban]

Hello, math!

Evidently, you don't know the "tabular method".

You have an excellent start!
I'll walk you through it the long way . . .

\(\displaystyle \L I \;= \;\int x^3\cdot e^x\,dx\)

. . \(\displaystyle \begin{array}{ccccccc}u &=& x^3 &\quad & dv & = & e^x\,dx \\ \\ \\
du & = &3x^2\,dx & \quad & v &=& e^x\end{array}\)


\(\displaystyle \L I \;= \;x^3\cdot e^x \,- \,\int e^x\cdot 3x^2\,dx\;\;\) Good!

We have: \(\displaystyle \L\:I \;= \;x^3\cdot e^x \,-\,3\int x^2\cdot e^x\,dx\)

We do "by parts" on this integral:

. . \(\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = &e^x\,dx \\ \\ \\
du & = &2x\,dx & \quad & v & = &e^x\end{array}\)

We have: \(\displaystyle \L\:I \;= \;x^3\cdot e^x \,-\,3\left[x^2\cdot e^x \,-\,\int e^x\cdot2x\,dx\right]\)

. . . . . . . .\(\displaystyle \L I \;=\;x^3\cdot e^x \,-\,3x^2\cdot e^x\,+\,6\int x\cdot e^x\,dx\)

And we do "by parts" on this integral, too:

. . \(\displaystyle \begin{array}{cccccc}u & = & x & \quad & dv & = &e^x\,dx \\ \\ \\ du & = & dx & \quad & v & = & e^x \end{array}\)

We have: \(\displaystyle \L\:I \;=\;x^3\cdot e^x \,-\,3x^2\cdot e^x \,+\,6\left[x\cdot e^x \,-\,\int e^x\,dx\right]\)

. . . . . . . .\(\displaystyle \L I \;=\;x^3\cdot e^x \,-\,3x^2\cdot e^x\,+\,6x\cdot e^x \,-\,6\int e^x\,dx\)

. . . . . . . .\(\displaystyle \L I \;=\;x^3\cdot e^x \,-\,3x^2\cdot e^x\,+\,6x\cdot e^x \,-\,6e^x\,+\,C\)


Answer: \(\displaystyle \L\:\:I \;= \;e^x\left(x^3\,-\,3x^2\,+\,6x\,-\,6\right) \,+\,C\)

 

[math]

thanks so much!!!

 

[Count Iblis]

You can also calculate the integral of Exp(p x) and then differentiate three times w.r.t. p.