Integral of a function with a square root in denominator

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The integral of the function (xdx)/sqrt(1-x^2) over the interval [2,5] was discussed, revealing a calculation error where the user incorrectly divided u^1/2 by 2 instead of 1/2. The correct integration yields 2√u, but there was confusion regarding the limits of integration after the variable change. The user initially calculated the integral as 0.716, but the correct value is 2.863 when properly evaluated. It was noted that the limits of integration must be adjusted accordingly after substitution. Overall, the discussion emphasized the importance of careful variable substitution and limit adjustments in integral calculus.
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Homework Statement


integrate (xdx)/sqrt(1-x^2) on the integreal [2,5]


Homework Equations


integral u^n(du) = u^n+1/(n+1)


The Attempt at a Solution


I'm no good at using latex so I scanned in my work.

I get .716 but when I check with my calculator it comes out as 2.863
 

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I see where I messed up. I was dividing u^1/2 by 2 instead of 1/2. Thanks anyways.
 


Two things:

\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.

Also you forgot to change the limits of integration after making the change of variables.
 


fzero said:
Two things:

\int u^{-1/2} du = \frac{u^{1/2}}{-\frac{1}{2}+1} = 2 \sqrt{u}.

Also you forgot to change the limits of integration after making the change of variables.

Actually I think I did do that. Not sure if u can see on the side where i did u(2) =5 and u(5) = 26.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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