Intergrating

[AAhmed]

Hi!

I am trying to do the following

Intergrate from x to 0. h(u)du= 4u^3/625-u^4 du

can anyone help?

 

[srmichael]

Hi!

I am trying to do the following

Intergrate from x to 0. h(u)du= 4u^3/625-u^4 du

can anyone help?

Well, that depends. Does \(\displaystyle h(u)=\dfrac{4u^3}{625}-u^4\) or \(\displaystyle h(u)=\dfrac{4u^3}{625-u^4}\)?

 

[AAhmed]

It's the later.

 

[srmichael]

It's the later.

OK. Now that we got that straight, what have you tried so far?

 

[AAhmed]

Not very much, I have not come across this type of intergration before. Powers above and below.

 

[pka]

Not very much, I have not come across this type of intergration before. Powers above and below.

What is the derivative of \(\displaystyle - \log (625 - {u^4})~?\)

 

[AAhmed]

My intergration is rusty.

Could it be exp(log(625-x^4)

 

[pka]

My intergration is rusty.

Could it be exp(log(625-x^4)

I did not ask any thing about integration. What is the derivative of the expression I posted?

 

[Quaid]

It's the later.

Oops, you typed the former.

pka gave you the verification step for an ansatz.



That is, if you recognize a form of the derivative of the bottom appearing on the top, then you can guess and check; sometimes, you may need to adjust a coefficient, but not in this exercise.

Cheers

 

[HallsofIvy]

Hi!

I am trying to do the following

Intergrate from x to 0. h(u)du= 4u^3/625-u^4 du

can anyone help?

So it is \(\displaystyle \int_0^x \frac{4u^3}{625- u^4}du\)

Substitute \(\displaystyle v= 625- u^4\)

 

[Quaid]

My intergration is rusty.

Oh. (We cross posted.)

Well, if you already had the antiderivative of the integrand, are you saying that you're not sure what to do with it (i.e., simplify the evaluated difference)?

I'm not sure where you're stuck or what you already know.