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/x^{1/2})dx
I got up to
(x^{2}/2)(x^{1/2}/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.
Integration
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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 
(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 
(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? 
Anyone able to do it?

(Original post by KingPanther)
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) 
(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/x^{1/2})dx is the original equation, although its just (x1/root x) which is the same as (x/1)(1/x^{1/2}) 
(Original post by KingPanther)
No, the answer is (10.2/3)
(x/1)(1/x^{1/2})dx is the original equation, although its just (x1/root x) which is the same as (x/1)(1/x^{1/2})
( 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 
(Original post by J.G.M)
Surely if the original equation was ( x1/root x ), this becomes ( x1/x^1/2 ), and that can be written as; x^1/2 (x1) 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
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 
(Original post by KingPanther)
I don't understand how you have written it, x(1/rootx) is the same as (x/1)(1/x^{1/2})
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
okay, so X1/root X is the same as X1/X^{1/2}, and the X^{1/2} can be bought up and made into X^{1/2}.
Therfore, X1/root X = X^{1/2} (X1), now you work out this multiplication to get; X^{1/2}  X^{1/2}
Now you integrate the equation; X^{1/2}  X^{1/2} dx
Which is ( X^{3/2} / 3/2 )  ( X^{1/2} / 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 
(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.... 

(Original post by just george)
Sounds like the book has had a mare and got it wrong then 
What would be the nth term for this sequence 3, 6, 11, 18, 27

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)
... 
n^2 +2

(Original post by J.G.M)
....(Original post by just george)
...(Original post by 0range)
...(Original post by f1mad)
...(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? (x1/x)^2 for example...... Also, how would I know what to sub into 4x^36x^2+2x when finding the turning points? 
(Original post by KingPanther)
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? (x1/x)^2 for example...... Also, how would I know what to sub into 4x^36x^2+2x when finding the turning points?
For your second question,
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 xcoordinates. Now substitute the values you will find for 'x' in the f(x) expression to find the ycoordinates.
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