# Solving an integral

this is the integral , and it's between (0,inf)
$$\int 4B^{3} x.exp(-2Bx).dx$$
how can i solve this?
it seems easy but i couldn't figure it out

I know that $$\int x^{n} exp(-x) dx =n!$$
i have to turn my integral into this format and solve the integral.so what to do next?

tiny-tim
Homework Helper
Welcome to PF!

I know that $$\int x^{n} exp(-x) dx =n!$$
i have to turn my integral into this format and solve the integral.so what to do next?

Hi PullMeOut! Welcome to PF!

You can also write that $$\int y^{n} exp(-y) dy =n!$$ …

so you need to put y = … ?

Maybe I can help but am not so literal in mathematics,its a vast language!What do you mean by exp(k)?Would that be 10^k?If so use INTEGRATION BY PARTS!When u have confirmed the above then if I have time I'l write down the step by step process!get me back quick!

I have found it
2Bx=y
dx2B=dy
then put this into the integral. it becomes like
$$\int \frac{4B^{3} y^{3}exp(-y) dy}{16B^{4}}$$
then it equals to $$\frac{3!}{4B}$$ with the help of the integral ı talk about in the previous post

well, i saw your message after i solved it, but thanks anyway tinytim , i really appreciate it :D

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Maybe I can help but am not so literal in mathematics,its a vast language!What do you mean by exp(k)?Would that be 10^k?If so use INTEGRATION BY PARTS!When u have confirmed the above then if I have time I'l write down the step by step process!get me back quick!
i think we can't use 10^k, becaouse it should be the same with dx , so it colud be x^k, but we solved the problem.you can check it out if you want and it is really an easy way.

tiny-tim
Homework Helper
i think we can't use 10^k, becaouse it should be the same with dx , so it colud be x^k, but we solved the problem.you can check it out if you want and it is really an easy way.

Hi PullMeOut!

Maybe I can help but am not so literal in mathematics,its a vast language!What do you mean by exp(k)?Would that be 10^k?If so use INTEGRATION BY PARTS!When u have confirmed the above then if I have time I'l write down the step by step process!get me back quick!

Hi natives!

Please don't try to help in homework threads if you don't understand one part of the question, and can't be bothered to read another part of the question.

exp(x) means ex.

And it is extremely unhelpful to tell an OP to use integration by parts when he has clearly stated "i have to turn my integral into this format and solve the integral"

Tiny tim,all I am trying to do is help and/or be helped,if I see other means I'd like to try them out,its a discussion forum not a class!And I dont think it is polite to tell him to ignore my suggestions,they are merely suggestions he/she can filter them out by themselves without external influence!Thank you.

Tiny tim,all I am trying to do is help and/or be helped,if I see other means I'd like to try them out,its a discussion forum not a class!
I think Tim agrees with this part.

And I dont think it is polite to tell him to ignore my suggestions,they are merely suggestions he/she can filter them out by themselves without external influence!Thank you.

In this case it is you've brought poor old OP on the wrong foot and you didn't read the question well enough.

I think Tim agrees with this part.

In this case it is you've brought poor old OP on the wrong foot and you didn't read the question well enough.

I wish you to have no further discussion over my actions about this,am good at what I do and I know what am doing..My bad I know so much and want to be clear with the alternatives at hand...Am sorry if my attempt was viewed wrong,bringing so thought "complex" mathematics to an OP..I thought its the passion to explore multiple possibilities that drive us all!Am alone.

HallsofIvy
Homework Helper
this is the integral , and it's between (0,inf)
$$\int 4B^{3} x.exp(-2Bx).dx$$
how can i solve this?
it seems easy but i couldn't figure it out
Looks to me like integration by parts, taking u= x, dv= exp(-2bx)dx would work nicely.

tiny-tim
Homework Helper
Looks to me like integration by parts, taking u= x, dv= exp(-2bx)dx would work nicely.

Hi HallsofIvy!

Yes, but in post #2, he says has has to use a different method.

I assume it's to provide practice at substitution.

(and , in principle, of course, substitution is easier and more natural than integration by parts)