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# Integration

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1. Integration
Hi, the brackets are not in the original equation, I've just added them to make it simpler.

The original equation is (x/1)-(1/x1/2)dx

I got up to

(x2/2)-(x1/2/1/2)

I'm pretty sure it is correct up until then, I'm unsure were to go from there... The numbers we sub in are 9 & 4 respectively.
Last edited by King-Panther; 01-04-2012 at 01:22.
2. Re: Integration
Sub 9 in first then sub 4 in
Then take the first answer where you subbed in 9 and take away the second answer that you got from subbing in 4
3. Re: Integration
(Original post by 0range)
Sub 9 in first then sub 4 in
Then take the first answer where you subbed in 9 and take away the second answer that you got from subbing in 4
No, thats no finished, it has to be simplified further... You can't have a fraction at the bottom of a fraction.... When I simplify the last fraction, I still dont get the right answer which is (10.2/3)
4. Re: Integration

substituting in 9 and 4:

So i guess must have gone wrong before this point if thats not the right answer :L

Are you sure the original equation is correct?
Last edited by just george; 01-04-2012 at 01:43.
5. Re: Integration
(Original post by just george)

substituting in 9 and 4:

So i guess must have gone wrong before this point if thats not the right answer :L

Are you sure the original equation is correct?
Yes, except in the original equation the second denominator is root x, which x1/2, thus (x/1)-(1/x1/2)dx, also its (x minus 1 over root x), which is the same as (x/1)-(1/x1/2)dx... I get the same answer as you, have you tried using the original equation?
Last edited by King-Panther; 01-04-2012 at 12:28.
6. Re: Integration
Anyone able to do it?
7. Re: Integration
(Original post by King-Panther)
No, thats no finished, it has to be simplified further... You can't have a fraction at the bottom of a fraction.... When I simplify the last fraction, I still dont get the right answer which is (10.2/3)
You don't have to simplify it, you won't get anymore marks unless it tells you to simplify your answer. Where I told you to sub it in is right. You may have copied the question down wrong or are looking at a different answer.
8. Re: Integration
(Original post by 0range)
You don't have to simplify it, you won't get anymore marks unless it tells you to simplify your answer. Where I told you to sub it in is right. You may have copied the question down wrong or are looking at a different answer.

(x/1)-(1/x1/2)dx is the original equation, although its just (x-1/root x) which is the same as (x/1)-(1/x1/2)
Last edited by King-Panther; 01-04-2012 at 12:27.
9. Re: Integration
(Original post by King-Panther)

(x/1)-(1/x1/2)dx is the original equation, although its just (x-1/root x) which is the same as (x/1)-(1/x1/2)
Surely if the original equation was ( x-1/root x ), this becomes ( x-1/x^1/2 ), and that can be written as; x^-1/2 (x-1) which equals [ x^1/2 - x^-1/2 ], which when integrated becomes;
( x^3/2 / 3/2 ) - ( x^1/2 / 1/2 )

However if what you say is the correct answer is right, this is also wrong (although i cant see how, sorry if there's any stupid mistakes) as i got the answer 32/3
10. Re: Integration
(Original post by J.G.M)
Surely if the original equation was ( x-1/root x ), this becomes ( x-1/x^1/2 ), and that can be written as; x^-1/2 (x-1) which equals [ x^1/2 - x^-1/2 ], which when integrated becomes;
( x^3/2 / 3/2 ) - ( x^1/2 / 1/2 )

However if what you say is the correct answer is right, this is also wrong (although i cant see how, sorry if there's any stupid mistakes) as i got the answer 32/3
I don't understand how you have written it, x-(1/rootx) is the same as (x/1)-(1/x1/2)

We both agree with that. I 'm lost with the rest, could you write how I have done please because 32/3 is the same as (10.2/3), thus the correct answer
Last edited by King-Panther; 01-04-2012 at 13:54.
11. Re: Integration

integrates to

substituting in 9 and 4:

if thats the right answer, then it is because you integrated instead of

edit: and writing 10.2/3 confused me.. that looks like whereas i think your meaning to say ?
Last edited by just george; 01-04-2012 at 14:42.
12. Re: Integration
(Original post by King-Panther)
I don't understand how you have written it, x-(1/rootx) is the same as (x/1)-(1/x1/2)

We both agree with that. I 'm lost with the rest, could you write how I have done please because 32/3 is the same as (10.2/3), thus the correct answer
ohhhh, sorry the 10.2/3 confused me, as it did with george, i thought you were writing it as a fraction, not a mixed number, but if you were meaning to say the answer's 102/3 then yeah, i guess my answer of 32/3 was right, woo! hah, okay i know george re-wrote it pretty clearly, but ill do so aswell just incase

okay, so X-1/root X is the same as X-1/X1/2, and the X1/2 can be bought up and made into X-1/2.

Therfore, X-1/root X = X-1/2 (X-1), now you work out this multiplication to get; X1/2 - X-1/2

Now you integrate the equation; X1/2 - X-1/2 dx

Which is ( X3/2 / 3/2 ) - ( X1/2 / 1/2 )

Now simply substitute 9 into the now integrated equation and work it out, then do the same with 4. Then minus the value you found while substituting 4 from the value you found from substituting 9. And voila! you get 32/3, which can also be writen as 102/3
Last edited by J.G.M; 01-04-2012 at 15:22. Reason: made clearer
13. Re: Integration
(Original post by just george)

integrates to

substituting in 9 and 4:

if thats the right answer, then it is because you integrated instead of

edit: and writing 10.2/3 confused me.. that looks like whereas i think your meaning to say ?
(Original post by J.G.M)
..

Is what is written in the book....
Last edited by King-Panther; 01-04-2012 at 16:07.
14. Re: Integration
(Original post by King-Panther)
Is what is written in the book....
Sounds like the book has had a mare and got it wrong then
15. Re: Integration
(Original post by just george)
Sounds like the book has had a mare and got it wrong then
Good, so i'm not wrong ..
16. Re: Integration
What would be the nth term for this sequence 3, 6, 11, 18, 27
17. Re: Integration
What would be the nth term for this sequence 3, 6, 11, 18, 27

(Original post by J.G.M)
....
(Original post by just george)
...
18. Re: Integration
n^2 +2
19. Re: Integration
(Original post by J.G.M)
....
(Original post by just george)
...
(Original post by 0range)
...
...
(Original post by raheem94)
...
(Original post by LeeM1)
n^2 +2

Thanks, I don't recall being taught finding the nth term in class... Maybe you can help me with the next problem, how would I find the second derivative of something? (x-1/x)^2 for example...... Also, how would I know what to sub into 4x^3-6x^2+2x when finding the turning points?
Last edited by King-Panther; 02-04-2012 at 22:17.
20. Re: Integration
(Original post by King-Panther)
Thanks, I don't recall being taught finding the nth term in class... Maybe you can help me with the next problem, how would I find the second derivative of something? (x-1/x)^2 for example...... Also, how would I know what to sub into 4x^3-6x^2+2x when finding the turning points?

Differentiate f(x), then make the dy/dx =0, because dy/dx is zero at turning points. So by solving it equal to zero you will get the x-coordinates. Now substitute the values you will find for 'x' in the f(x) expression to find the y-coordinates.

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Last updated: April 5, 2012
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