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# Integration watch

1. Edit: Okay nevermind I think there's nothing wrong with my method but I think I'm doing something wrong after.
After I've intergrated it I get -1/3(x+1)^3. This is a volumes of revolution question so it's actually (1/(x+1)^2)^2 . Then I have to find the area between 1 and 0 on along the x axis rotated 360 degrees. I put in the new intergral limits 2 and 1 but I don't seem to get the right answer.
2. Just make a substitution u=1+x or so
3. Have you covered integration by substitution?

4. My views on the subject of integration are that you should let whoever wants to wear a burka wear one. I know this is off-topic; I just came on this thread preparing to argue with bnp supporters. This thread is now about burkas.
5. (Original post by Necro Defain)
Just make a substitution u=1+x or so
Yes that's what I'm doing.
6. (Original post by Unknown?)
After I've intergrated it I get -1/3(x+1)^3.
Can you read my previous post? You shouldn't have a power of 3 or you need to add some more brackets because I can't follow what you are doing.
7. (Original post by Mr M)
Can you read my previous post? You shouldn't have a power of 3 or you need to add some more brackets because I can't follow what you are doing.
1 /3((x+1)^3)

sub u for x+1 so I have

1/u^4

make that

u^-4

and integrating that gives me the answer above.
8. (Original post by Mr M)

9. (Original post by Farhan.Hanif93)
10. (Original post by Mr M)
11. (Original post by Unknown?)
1 /3((x+1)^3)

sub u for x+1 so I have

1/u^4

make that

u^-4

and integrating that gives me the answer above.
You have lost a sign now. Anyway, assuming that bit is correct (LaTeX would really help as you have brackets in the wrong place), have you forgotten about pi?
12. (Original post by Farhan.Hanif93)
Don't think so. I have been using it for about 25 years fairly happily.
13. (Original post by Mr M)
Don't think so. I have been using it for about 25 years fairly happily.
If a=0, your above formula has a zero in the denominator!
All I'm saying is that there needs to be a condition up there stating that a must be non-zero. Most people will be able to spot it so I am being pretty pedantic but I guess there's not much wrong with mentioning that...
14. (Original post by Farhan.Hanif93)
If a=0, your above formula has a zero in the denominator!
All I'm saying is that there needs to be a condition up there stating that a must be non-zero. Most people will be able to spot it so I am being pretty pedantic but I guess there's not much wrong with mentioning that...
Sorry I see what you mean now. Thought you were saying n = 0. Yes, accepted.
15. (Original post by Mr M)
Sorry I see what you mean now. Thought you were saying n = 0. Yes, accepted.
I went a bit over the top with the pedantry there...
16. (Original post by Farhan.Hanif93)
I went a bit over the top with the pedantry there...
17. (Original post by Mr M)
Not gonna argue with that.

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