Integration via sub question (1 Viewer)

[zenger69]

I'm confused about the process...

For example in the question integrate x(3x-1)^4 using u=3x-1

now du=3 dx.

So what do you? Do you like multiply u by 3 or something?

And about if du = 1/2 dx

 

[magician]

I'm confused about the process...

For example in the question integrate x(3x-1)^4 using u=3x-1

now du=3 dx.

So what do you? Do you like multiply u by 3 or something?

And about if du = 1/2 dx

Can I just ask what ^ means? Lol.

 

[withoutaface]

You then say du/3=dx

and sub in /
............1/3| (u+1)u⁴du/3
................./

And then expand it.

 

[zenger69]

if it's all over 3 why do you need to multiply by 1/3?

 

[zenger69]

oh the symbol ^ means to the power of.

 

[Trev]

Can I just ask what ^ means? Lol.

ie. x^2 is x²

 

[Slidey]

Integrate I[x(3x-1)^4 dx] using u=3x-1

du/dx=3 --> du=3 dx
The integral is:

I[x(3x-1)^4 dx] = 1/3 * I[3 * x(3x-1)^4 dx] = I[(3x - 1 + 1)(3x-1)^4 dx]/3

We still can't substitute in du= 3 dx, so:

I[(3x - 1 + 1)(3x-1)^4]/3 = I[3(3x - 1 + 1)(3x-1)^4 dx]/9

Use the substitution, then:

I[(u+1)u^4 du]/9

Expand the integrand:

I[u^5+u^4 du]/9
=(u^6/6 + u^5/5)/9
=u^5(u/6+1/5)/9
=u^5(5u+6)/270

Substitute u=3x-1:
Final answer is:
(3x-1)^5(3x+1)/54 + C

I skip most of these steps normally - they're just there to show you.

 

[zergcave]

p.f?..........

 

[zenger69]

i'm good at 3U except i have no idea how to do it via substitution. So yeh... keep laughing..

 

[Euler]

by subsitution, the idea is to get rid of all the old variables (x) and make everything the new variable (u), and this applies also to the dx's and du's...

so if you have something like du=3dx, you want to find an expression for dx, like dx=1/3 du and then go back to the integral and substitute that in as well as the normal x's and u's.

also, don't forget to change the limits...