Heixi
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#1
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#1
I need to get ln…I didn’t …
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Heixi
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#2
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#2
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Should be 4+U^2 instead so 4+x
Last edited by Heixi; 1 month ago
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mqb2766
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#3
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#3
Check your integrand in u on the penultimate line. You know the result is ln(), so that must be a ....
Looks ok otherwise. I would have simply done dx/du = 2u though. Also remember its a definite integral, so you have limits.
Last edited by mqb2766; 1 month ago
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Heixi
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#4
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#4
(Original post by mqb2766)
Check your integrand in u on the penultimate line. You know the result is ln(), so that must be a ....
Looks ok otherwise. I would have simply done dx/du = 2u though. Also remember its a definite integral, so you have limits.
du/dx=1/2 x^-1
Last edited by Heixi; 1 month ago
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mqb2766
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#5
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#5
(Original post by Heixi)
1/2 x^-1
Not sure how you get that?
Just carefully apply your substitution to the original integral.
Last edited by mqb2766; 1 month ago
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Heixi
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#6
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#6
(Original post by mqb2766)
Not sure how you get that?
Just carefully apply your substitution to the original integral.
Oh so it should be 2f (2+u)^-1 du
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mqb2766
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#7
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#7
(Original post by Heixi)
Oh so it should be 2f (2+u)^-1 du
Yes, so just integrate.
Remember the original limits (definite integral) as well.
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Heixi
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#8
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#8
(Original post by mqb2766)
Yes, so just integrate.
Remember the original limits (definite integral) as well.
So 2ln|U+2|
Last edited by Heixi; 1 month ago
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mqb2766
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#9
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#9
(Original post by Heixi)
So 2ln|x+2|
Sort of, its "u" not "x" . Then apply the definite integral limits.
Note the two multiplier can brought inside the log as a power using the usual rules.
This gives something similar (not identical) to the expressions you give at the end of #2.
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Heixi
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#10
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#10
(Original post by mqb2766)
Sort of, its "u" not "x" . Then apply the definite integral limits.
Note the two multiplier can brought inside the log as a power using the usual rules.
This gives something similar (not identical) to the expressions you give at the end of #2.
ln|5+2 root 6 /2|
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mqb2766
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#11
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#11
(Original post by Heixi)
ln|5+2 root 6 /2|
Dont think so, but Ive no idea what youve done.
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Heixi
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#12
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#12
I’ve subbed the limits into u= x^1/2
Then ln(2+root 6)^2 - ln(4)
ln (( 2+root 6)^2)/4
The original limits are 36 ,0 not 4,2
Last edited by Heixi; 1 month ago
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mqb2766
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#13
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#13
(Original post by Heixi)
I’ve subbed the limits into u= x^1/2
Then ln(2+root 6)^2 - ln(4)
ln (( 2+root 6)^2)/4
Thats different from the previous expression and still not correct.
I still don't know how you get those terms.
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Heixi
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#14
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#14
(Original post by mqb2766)
Thats different from the previous expression and still not correct.
I still don't know how you get those terms.
Oh I think it because I subbed back the x terms instead of leaving it in U like I was supposed to
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mqb2766
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#15
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#15
(Original post by Heixi)
Oh I think it because I subbed back the x terms instead of leaving it in U like I was supposed to
Try and do it carefully and upload your full working if necessary?
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Heixi
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#16
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#16
(Original post by mqb2766)
Try and do it carefully and upload your full working if necessary?
I got ln16
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mqb2766
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#17
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#17
(Original post by Heixi)
I got ln16
Last try, can you upload your full working.

Being negative, you could guestimate the integral is ~0.3 as the width is 2 (4-2) and the height is ~1/6. So ln(16) is way out. However, it is part of the answer.
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#18
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#18
(Original post by mqb2766)
Last try, can you upload your full working.

Being negative, you could guestimate the integral is ~0.3 as the width is 2 (4-2) and the height is ~1/6. So ln(16) is way out. However, it is part of the answer.
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mqb2766
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#19
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#19
(Original post by Heixi)
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So the limits are different from the OP? They were 2 to 4, now they're 0 to 36? ln(16) is correct for these limits.
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Heixi
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#20
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#20
(Original post by mqb2766)
So the limits are different from the OP? They were 2 to 4, now they're 0 to 36? ln(16) is correct for these limits.
It was written wrong ,sorry .
Thanks so much for the help
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