I can't do this question :-

Integrate the function f(x) = (x^3.e^-x^2)dx

I have tried many techniques but none seem to be working. can you please also give links of some useful techniques for questions such as these?

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- Jan 3rd 2008, 06:05 AMAltairHelp Required !!
I can't do this question :-

Integrate the function f(x) = (x^3.e^-x^2)dx

I have tried many techniques but none seem to be working. can you please also give links of some useful techniques for questions such as these? - Jan 3rd 2008, 06:10 AMKrizalid
Have you tried settin' $\displaystyle u=-x^2$?

- Jan 3rd 2008, 06:14 AMcolby2152
We have an integrand of: $\displaystyle x^3e^{-x^2}$

Use u-substitution by setting $\displaystyle u = x^2$

$\displaystyle du = 2xdx$

Now, the integrand looks like: $\displaystyle \frac{u}{2}e^{-u}$

Use integration by parts or tabular integration for this to get a result of: $\displaystyle \frac{-e^{-u}}{2}(u+1)$... subbing $\displaystyle x^2 = u$ takes us to a final answer of: $\displaystyle \frac{-e^{-x^2}}{2}(x^2+1)$ - Jan 5th 2008, 06:14 AMAltair
Thank you.

Can you please do the integration by parts step-by-step (in terms of U) for me as well? Cuz I seem to be making some mistake in it.

Thank You - Jan 5th 2008, 08:17 AMSoroban
Hello, Altair!

Quote:

$\displaystyle \int x^3e^{-x^2}\,dx$

. . $\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = & xe^{-x^2}dx \\ du & = & 2x\,dx & \quad & v & = & -\frac{1}{2}e^{-x^2} \end{array}$

We have: .$\displaystyle -\frac{1}{2}x^2e^{-x^2} + \int xe^{-x^2}dx$

Hence: .$\displaystyle -\frac{1}{2}x^2e^{-x^2} - \frac{1}{2}e^{-x^2} + C \;=\;-\frac{1}{2}e^{-x^2}\left(x^2+1\right) + C$

- Jan 5th 2008, 09:45 AMdtomphysics
A roller coaster at an amusement park starts out on a level track 50.0 m long and then goes up a 25.0-m incline at an angle of 30 degree to the horizontal. It then goes down a 15.0-m ramp with an incline of 40 degree to the horizontal. When the roller coaster has reached the bottom of the ramp, what is its displacement from its starting point?(Handshake)

- Jan 5th 2008, 12:02 PMgalactus
I would suggest a new thread for a new problem. Folks are more likely to see it that way.

The displacement is how far it traveled from one point to another regardless of the 'hills and valleys'.

That's why it's called diplacement. How much distance it displaces.

It traveled horizontally for 50 feet. Then, using the law of cosines, it traveled

$\displaystyle x=25cos(30)$ and then $\displaystyle y=15cos(40)$

$\displaystyle 50+25cos(30)+15cos(40)$