someone plz check my solutions.. (1 Viewer)

[ToO LaZy ^*]

question 1...

 

[ToO LaZy ^*]

solution?

 

[ToO LaZy ^*]

q2

 

[ToO LaZy ^*]

q2 solution?

 

[CM_Tutor]

Solution of question 1 is wrong (Sorry). Error in going between first and second lines. You have written

int (from 0 to 2) sqrt(x / (4 - x)) dx = int (from 0 to 2) x / sqrt(4 - x²) dx

This involves multiplying numerator by sqrt(x) and denominator by something else. You mean:

int (from 0 to 2) sqrt(x / (4 - x)) dx = int (from 0 to 2) x / sqrt(4x - x²) dx

 

[CM_Tutor]

Question 2 is fine.

 

[ToO LaZy ^*]

Originally posted by CM_Tutor
Solution of question 1 is wrong (Sorry). Error in going between first and second lines. You have written

int (from 0 to 2) sqrt(x / (4 - x)) dx = int (from 0 to 2) x / sqrt(4 - x²) dx

This involves multiplying numerator by sqrt(x) and denominator by something else. You mean:

int (from 0 to 2) sqrt(x / (4 - x)) dx = int (from 0 to 2) x / sqrt(4x - x²) dx

ohhhk..i do it again.

 

[ToO LaZy ^*]

Originally posted by CM_Tutor
Question 2 is fine.

yay
i'll probably have more for you to check later on..so stayed tuned..(plz).

 

[ToO LaZy ^*]

this is question 1 again..

 

[CM_Tutor]

Closer, but still not right. There are two problems:

1. (2x - 4) / sqrt(4x - x²) should not integrate to a log.

2. When you integrated to get the inverse sine, you have not taken account of the fact that there is a minus sign in front of the x.

NOTE: The function sqrt(x / (4 - x)) is non-negative throughout 0 <= x <= 2, so the answer is positive. Furthermore, since the max value of this function is 1 on this domain, it's not hard to show that the answer must be between 1 and 2. You should always do a quick check like this - reflect on an answer by asking 'is my answer reasonable?'.

PS: I get the answer to be pi - 2, and I won't be around for about the next hour.

 

[ToO LaZy ^*]

dag nabit!...thx CM
can someone help me out..i've lost all self esteem.

 

[CM_Tutor]

int (4 - 2x) / sqrt(4x - x²) dx = 2 * sqrt(4x - x²) + C, for some constant C

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

 

[kimmeh]

me me me i wanna do it brb doing it

 

[ToO LaZy ^*]

Originally posted by CM_Tutor
int (4 - 2x) / sqrt(4x - x²) dx = 2 * sqrt(4x - x²) + C, for some constant C

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

sorry...but...how did that happen?

 

[ToO LaZy ^*]

Originally posted by kimmeh
me me me i wanna do it brb doing it

yay!

 

[kimmeh]

too lazy, u made ur error in line 6. The first intergral is not a log

 

[ToO LaZy ^*]

what i don;t get is the circled bit..
is there some rule that i may have missed?

 

[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

 

[kimmeh]

its like a quick thing, not a rule-if you remeber is, it will make integrating those functions faster

 

[kimmeh]

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

i'm right