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ok..now i understand that bit..Originally posted by ToO LaZy ^*
what i don;t get is the circled bit..
Originally posted by kimmeh
i'm right![]()
I GET IT NOW!..EXCUSE MY SCREAMING..BUT I'M JUST TOO HAPPYOriginally posted by CM_Tutor
int 1 / sqrt(4x - x<sup>2</sup>) dx = int 1 / sqrt[4 - (2 - x)<sup>2</sup> dx = - sin<sup>-1</sup>[(2 - x) / 2] + C, for some constant C
Actually, both 4 - (2 - x)<sup>2</sup> and 4 - (x - 2)<sup>2</sup> are correct.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
Apart from the fact that the question was a definite integral, and you left the inverse sine part as an indefinite integral.Originally posted by kimmeh
i'm right![]()
It might be easier to follow if you break it up into steps and do the substitutions:Originally posted by Teoh
I so have no idea how this question has been solved...
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 - x<sup>2</sup>) dx
Complete square to int (from 0 to 2) x / sqrt(4 - (2 - x)<sup>2</sup>) dx
Use u = 2 - x to convert to int (from 2 to 0) (2 - u) / sqrt(4 - u<sup>2</sup>) * -1 du
which is int (from 0 to 2) (2 - u) / sqrt(4 - u<sup>2</sup>) du
Split into two as: 2 * int (from 0 to 2) 1 / sqrt(4 - u<sup>2</sup>) du + int (from 0 to 2) -u / sqrt(4 - u<sup>2</sup>) du
The first of these is 2 * [sin<sup>-1</sup>(u / 2)] (from 0 to 2), which is 2 * sin<sup>-1</sup>(1) = pi
The second is (1 / 2) * int (from 0 to 2) -2u / sqrt(4 - u<sup>2</sup>) du
This is (1 / 2) * [2 * sqrt(4 - u<sup>2</sup>)] (from 0 to 2), which is -2 (this can be
i would think otherwiseOriginally posted by ToO LaZy ^*
i think i'll use your method...less complicated
can somone confirm that it's right/wrong?..i've got this question on my test coming up..Originally posted by ToO LaZy ^*
one more to check plz
is the answer to part one of the question sufficient?
I don't know if it's too late, but the solution is insufficient (sorry).Originally posted by ToO LaZy ^*
can somone confirm that it's right/wrong?..i've got this question on my test coming up..