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# C4 Integration watch

1. Hi. Just doing Maths and come across this question with this mark scheme (OCR MEI C4 Jan 06)
I am wondering how comes it is 1/2lny-1/2ln(2-y) instead of 1/2lny+1/2ln(2-y)?
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2. (Original post by prepdream)
Hi. Just doing Maths and come across this question with this mark scheme (OCR MEI C4 Jan 06)
I am wondering how comes it is 1/2lny-1/2ln(2-y) instead of 1/2lny+1/2ln(2-y)?
Integrate 1/[2(2-y)]. There's a minus in there.

Differentiating 1/2 * ln (2 - y) creates a minus sign at the front.
3. (Original post by ombtom)
Integrate 1/[2(2-y)]. There's a minus in there.

Differentiating 1/2 * ln (2 - y) creates a minus sign at the front.
Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw
4. (Original post by prepdream)
Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw
You're right that it's the same as 1/2 * integral of 1/(2-y).

Integrating 1/(2-y) gives -ln(2-y).

Differentiating ln(2-y) gives -1/(2-y), hence the need for a minus sign.

Hope this helps.
5. (Original post by prepdream)
Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw
The integral of is .
6. (Original post by prepdream)
Sorry, still don't get it By integrating 1/[2(2-y)], shouldn't it be 1/2*integration of 1/(2-y), which means 1/2ln(2-y) without a minus? :/
Thanks for the reply btw
We know that , yeah? That is, if the numerator is the derivative of the denominator, then integrating it produces the logarithm of the denominator.

In this case, we have: - the numerator is almost the derivative of the denominator - the dervative of the denominator is

How do we fix this? Well,

since we now have the numerator as the derivative of the numerator.

Try this out on: - the numerator is almost the derivative of the numerator.

If you can't spot what to do:
Spoiler:
Show
7. (Original post by Zacken)
x
This doesn't look much like Physics.
8. (Original post by tinkerbella~)
This doesn't look much like Physics.
Physics is maths basically
9. (Original post by Zacken)
We know that , yeah? That is, if the numerator is the derivative of the denominator, then integrating it produces the logarithm of the denominator.

In this case, we have: - the numerator is almost the derivative of the denominator - the dervative of the denominator is

How do we fix this? Well,

since we now have the numerator as the derivative of the numerator.

Try this out on: - the numerator is almost the derivative of the numerator.

If you can't spot what to do:
Spoiler:
Show
Ahah alright that makes so much sense. I completely forgot about the f'(x)/f(x) thing. Thank you!
10. (Original post by prepdream)
Ahah alright that makes so much sense. I completely forgot about the f'(x)/f(x) thing. Thank you!
No problem! Do you want to test your understanding on the problem I gave at the end?
11. (Original post by Zacken)
No problem! Do you want to test your understanding on the problem I gave at the end?
Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
12. (Original post by prepdream)
Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
Yes!
13. (Original post by prepdream)
Yeah is that -1/2ln(2-x^2) if I am not wrong? :/
Well technically it should be
.

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