jumblehunter
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#1
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#1
Apologies for another post, the forum is kinda snowed under with jumblehunter posts today ( see what I did there?)

Anyway, I need to integrate y=4x(2x-1)^7 with the substitution 2x-1 and bounds 0.5 and 0

u=2x-1
du/dx=2
du/2=dx

Thus I got the original equation simplified to *integral sign with bounds 0 and -1* 2x(u)^7

I then simplified this to
*integral sign with bounds 0 and -1* (u+1)(u)^7

I don't know where to go from here. Is this right so far and can I have some help please?
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L'Evil Fish
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#2
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Assuming your limits are right,expand the brackets

You should be at:

\displaystyle\int_{0}^{-1} (u+1)(u^7) du
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steve2005
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#3
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(Original post by jumblehunter)
Apologies for another post, the forum is kinda snowed under with jumblehunter posts today ( see what I did there?)

Anyway, I need to integrate y=4x(2x-1)^7 with the substitution 2x-1 and bounds 0.5 and 0

u=2x-1
du/dx=2
du/2=dx

Thus I got the original equation simplified to *integral sign with bounds 0 and -1* 2x(u)^7

I then simplified this to
*integral sign with bounds 0 and -1* (u+1)(u)^7

I don't know where to go from here. Is this right so far and can I have some help please?
Expand the bracket.
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jumblehunter
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#4
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(Original post by L'Evil Fish)
Assuming your limits are right,expand the brackets

You should be at:

\displaystyle\int_{0}^{-1} (u+1)(u^7) du
yeah thats where I'm at, excuse my poor notation.
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jumblehunter
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#5
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(Original post by jumblehunter)
yeah thats where I'm at, excuse my poor notation.
Also, why does the -1 go above the zero on the integral?
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steve2005
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#6
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(Original post by L'Evil Fish)
Assuming your limits are right,expand the brackets

You should be at:

\displaystyle\int_{0}^{-1} (u+1)(u^7) du
I think you have the limits the wrong way round.
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L'Evil Fish
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(Original post by steve2005)
I think you have the limits the wrong way round.
(Original post by jumblehunter)
yeah thats where I'm at, excuse my poor notation.
(Original post by jumblehunter)
Also, why does the -1 go above the zero on the integral?
Well if the original one was:

\int_{\frac{1}{2}}^{0} then it'd be what I said, but I think you meant the limits are \int_{0}^{\frac{1}{2}} in which case the new limits will be int_{-1}^{0} however it's not always the case that the smaller number is under!
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jumblehunter
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#8
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Okay I'll expand this out with new limits and see what I get
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L'Evil Fish
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#9
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(Original post by jumblehunter)
Okay I'll expand this out with new limits and see what I get
Okay!
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