someone plz check my solutions.. (1 Viewer)

[Teoh]

I so have no idea how this question has been solved...

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*
what i don;t get is the circled bit..

ok..now i understand that bit..

but don't know what happened in the next step


-sqrt (4x-x^2)

 

[Teoh]

Complete the square to use inverse sin/cos rule of integration.

Your previous post about the completing square can't be right
Cause you'd be left with 4x, rather then -4x

 

[ToO LaZy ^*]

Originally posted by kimmeh
i'm right

Originally posted by CM_Tutor


int 1 / sqrt(4x - x²) dx = int 1 / sqrt[4 - (2 - x)² dx = - sin^-1[(2 - x) / 2] + C, for some constant C

I GET IT NOW!..EXCUSE MY SCREAMING..BUT I'M JUST TOO HAPPY
THX CM AND KIMMEH...

 

[kimmeh]

ToO LaZy ^* , PM'd you


ahahaha no problemoes

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
and i think there was a mistake in the completing of the square bit.
it should be: 4 - (2-x)^2....i think

Actually, both 4 - (2 - x)² and 4 - (x - 2)² are correct.

Originally posted by kimmeh
i'm right

Apart from the fact that the question was a definite integral, and you left the inverse sine part as an indefinite integral.

And the answer is pi - 2.

Originally posted by Teoh
I so have no idea how this question has been solved...

It might be easier to follow if you break it up into steps and do the substitutions:

Start with int (from 0 to 2) sqrt(x / (4 - x)) dx
Convert to int (from 0 to 2) x / sqrt(4x - x²) dx
Complete square to int (from 0 to 2) x / sqrt(4 - (2 - x)²) dx
Use u = 2 - x to convert to int (from 2 to 0) (2 - u) / sqrt(4 - u²) * -1 du
which is int (from 0 to 2) (2 - u) / sqrt(4 - u²) du
Split into two as: 2 * int (from 0 to 2) 1 / sqrt(4 - u²) du + int (from 0 to 2) -u / sqrt(4 - u²) du
The first of these is 2 * [sin^-1(u / 2)] (from 0 to 2), which is 2 * sin^-1(1) = pi
The second is (1 / 2) * int (from 0 to 2) -2u / sqrt(4 - u²) du
This is (1 / 2) * [2 * sqrt(4 - u²)] (from 0 to 2), which is -2 (this can be confirmed by a v = 4 - u² substitution, if necessary).

Thus, the answer is pi - 2.

 

[ToO LaZy ^*]

wow thanks..you've cleared everything up for me
i think i'll use your method...less complicated

 

[ToO LaZy ^*]

one more to check plz

is the answer to part one of the question sufficient?

 

[ToO LaZy ^*]

?

 

[ToO LaZy ^*]

don't worry...i found out

 

[kimmeh]

Originally posted by CM_Tutor
Start with int (from 0 to 2) sqrt(x / (4 - x)) dx
Convert to int (from 0 to 2) x / sqrt(4x - x²) dx
Complete square to int (from 0 to 2) x / sqrt(4 - (2 - x)²) dx
Use u = 2 - x to convert to int (from 2 to 0) (2 - u) / sqrt(4 - u²) * -1 du
which is int (from 0 to 2) (2 - u) / sqrt(4 - u²) du
Split into two as: 2 * int (from 0 to 2) 1 / sqrt(4 - u²) du + int (from 0 to 2) -u / sqrt(4 - u²) du
The first of these is 2 * [sin^-1(u / 2)] (from 0 to 2), which is 2 * sin^-1(1) = pi
The second is (1 / 2) * int (from 0 to 2) -2u / sqrt(4 - u²) du
This is (1 / 2) * [2 * sqrt(4 - u²)] (from 0 to 2), which is -2 (this can be

Originally posted by ToO LaZy ^*
i think i'll use your method...less complicated

i would think otherwise

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*
one more to check plz

is the answer to part one of the question sufficient?

can somone confirm that it's right/wrong?..i've got this question on my test coming up..

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
can somone confirm that it's right/wrong?..i've got this question on my test coming up..

I don't know if it's too late, but the solution is insufficient (sorry).

You have shown that [sin(2n + 1)θ - sin(2n - 1)θ] / sin θ = 2cos 2nθ

What you then need to say is that I_n - I_n-1 = ∫ (from 0 to π / 2) [sin(2n + 1)θ - sin(2n - 1)θ] / sin θ dθ
= ∫ (from 0 to π / 2) 2cos 2nθ dθ
= 2[(1 / 2n) * sin 2nθ] (from 0 to π / 2)
= (1 / n) * [sin(2nπ / 2) - sin 0]
= (1 / n) * [sin nπ - sin 0]
= (1 / n) * [0 - 0]
= 0, as required.

Your argument from cos 2nθ to 0 is not clear.

 

[ToO LaZy ^*]

yeah, it is a bit late, but it's cool, coz i figured i had to do the extra bit- the bit you just showed me
i'm glad to say i got 100% for my integration assessment task ..thx CM