# Integration by parts/substitution

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Hi. I think I jumped too eagerly into my latest module, as I am positive it makes many assumptions on previous experience. I am obviously lacking.

I have created a question in Word, but the equations don;t appear to be copying into this thread. So I have uploaded question instead. If this is against the rules please let me know, and I will attempt to re-create the equations here instead.

The question involves integration of ∫x(^(a-1)) e(^(-bx)) dx - and I have shown in the word doc what I think I need to do. However, the final answer is:

Γ(a)/ba

Which I know I won;t get from integration, but my answer should relate to this answer, using one of the distribution methods.

Many thanks

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BvU
Homework Helper
Hi,

Make life easy for yourself and study how to apply partial integration on ##\displaystyle \int_0^\infty x^n e^{-x} dx##

(and yes, the Gamma function is involved)

Nick Jarvis
Thanks BvU, will take a look this evening.

ok. I have integrated ∫xn ex using parts and I get:

xn ex - ∫exnxn-1 this equates (I hope) to:

ex(xn - nxn-1) this can be simplified obvs. Is this correct?

Thanks

However, if I have ∫BxB-1 e-xB dx, can I not integrate by substitution?

Let u = xB so du = BxB-1dx which gives:

∫e-udu and therefore the answer is -e-xB + c

I know that is not correct, but I thought I could integrate by either substitution OR parts? This integration is key to me understanding stats/probability, but it has been many years since I have integrated.

Just looked again, and I know that ∫BxB-1 = xB, but how do I incorporate that into it when I also have exB

Many thanks

BvU
Homework Helper
and I get: xn ex - ∫exnxn-1 dx
correct so far.
this equates (I hope) to: ex(xn - nxn-1)
Your hope is in vain. You can not just make an ##\displaystyle\int## sign disappear !

Do not confuse yourself by omitting the ##dx##.

What are the bounds of the integral ? What did you do with them ? What is the meaning of the (correct) expression you got from integration by parts ? ##\quad##  of the wrong expression ##xe^x## instead of ##xe^{-x}##

Nick Jarvis
BvU