Quick Q on By Parts
Maths and statistics discussion, revision, exam and homework help.
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Quick Q on By PartsMy question is that when you have by parts with limits, the how do you lay it out ? I can't find examples of this anywhere in my textbooks. Because we set up an intermediary form which has another integral in it, what limits does that have ?
Which one of the following is the proper way of laying this out, feel free to post the proper way if it is none of the following.
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow x\ln x - \int^a_b 1\ dx
\
\Rightarrow \left[ x\ln \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/ea/ead4b7274a0db0cd3c68f0298e3db4b5.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow x\ln x - \int^a_b 1\ dx
\
\Rightarrow \left[ x\ln \ - x \right]_b^a")
Or
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x - \int 1\ dx \right]_b^a
\
\Rightarrow \left[ x\ln \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/06/06a15d9a99acd8c7f03b750e3b19cc0b.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x - \int 1\ dx \right]_b^a
\
\Rightarrow \left[ x\ln \ - x \right]_b^a")
Or
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x - \int^a_b 1\ dx \right]_b^a
\
\Rightarrow \left[ x\ln \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/17/17c3c09c912bb31097caeb40bef7cb5a.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x - \int^a_b 1\ dx \right]_b^a
\
\Rightarrow \left[ x\ln \ - x \right]_b^a")
ThnxLast edited by member910132; 11-05-2012 at 09:05. -
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Re: Quick Q on By PartsTo my mind:(Original post by member910132)
My question is that when you have by parts with limits, the how do you lay it out ? I can't find examples of this anywhere in my textbooks. Because we set up an intermediary form which has another integral in it, what limits does that have ?
Which one of the following is the proper way of laying this out, feel free to post the proper way if it is none of the following.
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow x\ln x - \int^a_b 1\ dx
\
\Rightarrow \left[ x\ln x \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/c4/c41c3d43eaa215d7c3a0d967b02f8d94.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow x\ln x - \int^a_b 1\ dx
\
\Rightarrow \left[ x\ln x \ - x \right]_b^a")
Or
Or
Thnx
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x]_b^a - \int^a_b 1\ dx \right
\Rightarrow \left[ x\ln x \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/7d/7d7afb1488981d680c215379fb1d1b57.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x]_b^a - \int^a_b 1\ dx \right
\Rightarrow \left[ x\ln x \ - x \right]_b^a")
Last edited by Slumpy; 11-05-2012 at 08:34. -
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Re: Quick Q on By PartsThe one that makes the most sense is the first one but they're pretty easy with letting you off on this kind of thing. Except for personally; I'd write it as(Original post by member910132)
My question is that when you have by parts with limits, the how do you lay it out ? I can't find examples of this anywhere in my textbooks. Because we set up an intermediary form which has another integral in it, what limits does that have ?
Which one of the following is the proper way of laying this out, feel free to post the proper way if it is none of the following.
Or
Or
Thnx
![[xln x]^a_b - \int^a_b 1 dx](http://www.thestudentroom.co.uk/latexrender/pictures/e4/e4aae5bc8b30e03bcb0285581cd73466.png "[xln x]^a_b - \int^a_b 1 dx")
And I think that's the 'most correct' way of doing the first step.
Also those implies signs are used incorrectly. You aren't implying anything - you just need a regular equals sign.
EDIT: You've also missed the 'x' in 'lnx' a lot
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Re: Quick Q on By PartsEven in STEP are they easy about that ?(Original post by hassi94)
The one that makes the most sense is the first one but they're pretty easy with letting you off on this kind of thing. Except for personally; I'd write it as
And I think that's the 'most correct' way of doing the first step.
Also those implies signs are used incorrectly. You aren't implying anything - you just need a regular equals sign.
EDIT: You've also missed the 'x' in 'lnx' a lot
They are \Rightarrows, but yea "=" should be used.
I just missed 1 "x" in ln x, the rest was a copy paste job lol -
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Re: Quick Q on By Parts\Rightarrow is an implication.(Original post by member910132)
Even in STEP are they easy about that ?
They are \Rightarrows, but yea "=" should be used.
I just missed 1 "x" in ln x, the rest was a copy paste job lol -
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Re: Quick Q on By Parts
So you guys are saying that(Original post by hassi94)
The one that makes the most sense is the first one but they're pretty easy with letting you off on this kind of thing. Except for personally; I'd write it as
And I think that's the 'most correct' way of doing the first step.
Also those implies signs are used incorrectly. You aren't implying anything - you just need a regular equals sign.
EDIT: You've also missed the 'x' in 'lnx' a lot
![\displaystyle \left[F(x) \right]_b^a - \left[G(x) \right]_b^a = \left[F(x) - G(x) \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/fa/fa95ace25166f25877fdee3a2d711821.png "\displaystyle \left[F(x) \right]_b^a - \left[G(x) \right]_b^a = \left[F(x) - G(x) \right]_b^a")
?
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Re: Quick Q on By PartsIn this case the Q is actually(Original post by tamimi)
Int by parts is probably my least favourite topic in A level maths.
Just can't work my head around it.
but we put the 1 in there to make it easier. u=ln x v'= 1 and proceed from there.
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Re: Quick Q on By PartsAlso yes from what I understand even in STEP they are fairly indifferent about it. Especially with subsitutions - sometimes it's not worth changing the limits of your integral (and just changing the results back to the original variable and subbing the original limits in is easier) so they don't mind if you have limits - stop having limits then have them again(Original post by member910132)
Even in STEP are they easy about that ?
They are \Rightarrows, but yea "=" should be used.
I just missed 1 "x" in ln x, the rest was a copy paste job lol As long as you don't completely lose them for whatever reason.
![\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x]_b^a - \int^a_b 1\ dx \right
\Rightarrow \left[ x\ln x \ - x \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/bb/bb33ab54e087534cd6513e30cbd39040.png "\displaystyle \int^a_b (1)\times \ln x\ dx
\
\Rightarrow \left[x\ln x]_b^a - \int^a_b 1\ dx \right
\Rightarrow \left[ x\ln x \ - x \right]_b^a")
![\displaystyle \left[F(x) \right]_b^a = F(a) - F(b)](http://www.thestudentroom.co.uk/latexrender/pictures/1d/1dda68e5a3e11af6866c81508589ea9a.png "\displaystyle \left[F(x) \right]_b^a = F(a) - F(b)")
![\left[G(x) \right]_b^a = G(a) - G(b)](http://www.thestudentroom.co.uk/latexrender/pictures/7e/7ec07b0e7e5ae455f96c23634da500ba.png "\left[G(x) \right]_b^a = G(a) - G(b)")
![\displaystyle \left[F(x) \right]_b^a - \left[G(x) \right]_b^a = [F(a) - F(b)] - [G(a) - G(b)]](http://www.thestudentroom.co.uk/latexrender/pictures/94/94e761583ec0a55fb249a7f07ae2f6c5.png "\displaystyle \left[F(x) \right]_b^a - \left[G(x) \right]_b^a = [F(a) - F(b)] - [G(a) - G(b)]")
![= [F(a) - G(a)] - [F(b) - G(b)] = \left[F(x) - G(x) \right]_b^a](http://www.thestudentroom.co.uk/latexrender/pictures/01/0178f9975554b7086ea5a9828bdd58c9.png "= [F(a) - G(a)] - [F(b) - G(b)] = \left[F(x) - G(x) \right]_b^a")
