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Reply 20
This isn't my work, but I think its a useful point made by NonCommAlg in a different forum (where Tom has posted the same question)


Hint: let f(x)=x+22x4+x22x4,f(x)=\sqrt {x + 2\sqrt {2x - 4}} + \sqrt {x - 2\sqrt {2x - 4}}, then: (f(x))2=2x+2x4.(f(x))^2=2x+2|x-4|. thus: f(x)={22if  3x42x2if  4x5.f(x)=\begin{cases} 2\sqrt{2} & \text{if} \ \ 3 \leq x \leq 4 \\ 2 \sqrt{x-2} & \text{if} \ \ 4\leq x \leq 5 \end{cases}.
Alex,

You may be delighted to see I integrated it using your substiution and obtained answer (B) as well. The correct answer is still (E) however! I'm not clear what the problem is just yet.

However, looking at the graph, there is a problem at x = 4.
Reply 22
SimonM
This isn't my work, but I think its a useful point made by NonCommAlg in a different forum (where Tom has posted the same question)


That's what I was trying to figure out. My sketch was bad though.
heres what i got, but from what other people are posting im doing it all wrong D:

Spoiler

Reply 24
Mr M
I'm not clear what the problem is just yet.


Pick me pick me!

I think you have missed some modulus signs taking a certain square root

u2+2+2u=(u2+2)u2+2\sqrt{\frac{u}{2}+2+2\sqrt{u}} = |(\frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2})| \not= \frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2}
Reply 25
NonCommAlg's solution certainly makes sense.

My only worry was that it is usually unlikely for "None of these" to be the answer to a question.
Reply 26
alexmahone

My only worry was that it is usually unlikely for "None of these" to be the answer to a question.


Heh, its got to happen at some point otherwise that would be a waste of a choice.
Reply 27
SimonM
Heh, its got to happen at some point otherwise that would be a waste of a choice.


That settles the matter then.
Reply 28
SimonM

u2+2+2u=(u2+2)u2+2\sqrt{\frac{u}{2}+2+2\sqrt{u}} = |(\frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2})| \not= \frac{\sqrt{u}}{\sqrt{2}}+\sqrt{2}


Why not for 2 <= u <= 6?
Reply 29
Either that or the problem setters made a mistake
Reply 30
Jeqle
Why not for 2 <= u <= 6?


Good spot, its the other bracket which was the problem, but I'd already typed that one out and couldn't be bothered with the other
Reply 31
So just to clarify, you're saying when we consider

26(u2)2du\int^6_2 \sqrt{(\sqrt{u} - 2)^2} \mathrm{d}u

We have to say

=242udu+46u2du= \int^4_2 2 - \sqrt{u} \mathrm{d}u + \int^6_4 \sqrt{u} - 2 \mathrm{d}u

Am I right?
Reply 32
Yes
so is it

Spoiler



i hope so...
Chaoslord
so is it

Spoiler



i hope so...


No that one is REALLY wrong.

You have forgotten to multiply by u when replacing dx by u du.

And you still have the same problem when x = 4.
And there you go thomas .. you weren't ignored were you?
Mr M
No that one is REALLY wrong.

You have forgotten to multiply by u when replacing dx by u du.

And you still have the same problem when x = 4.


okay i changed it silly mistake sorry

i must be really ignorant when x = 4 my u = 2 which i can't see a problem with, give me a min to do my final subs im guessing thats where i go wrong
Chaoslord


26u(12(u+2)2+12(u2)2)du\displaystyle\int^{\sqrt{6}}_{\sqrt{2}} u(\sqrt{\frac{1}{2}(u+2)^2} + \sqrt{\frac{1}{2} (u - 2)^2})\, du



Your problem is with the second term in this line. Think about when u<2u < 2.
Mr M
Your problem is with the second term in this line. Think about when u<2u < 2.


but dont we have to integrate before we think about substituting the u?

edit: if u < 2 cause im squaring it its still +ve?
Reply 39
Chaoslord
but dont we have to integrate before we think about substituting the u?

edit: if u < 2 cause im squaring it its still +ve?


You didn't do that in your above post, or note that the sqaure root of a function whose square has been completed is the modulus. The integral shown a few lines above in your working is not valid for u < 2.

If you are unsure, see the 1959 IO question I posted up a few days ago. It is exactly the same as this (without the integrating), but you certainly get an idea of where the modulus function comes in.

See simons working at the top of this thread.