How to Solve the Integral 2x((1-x^2)^(1/2)) and Evaluate at x=1

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The integral 2x((1-x^2)^(1/2)) presents challenges in solving, particularly when evaluating at x=1, where the numerator results in zero. Attempts at integration by parts and substitution have not yielded success. A suggestion is made to use the substitution u=1-x^2, which may simplify the problem. The discussion emphasizes that having a zero numerator is not problematic unless it occurs in a denominator. The conversation revolves around finding an effective method to solve the integral.
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Homework Statement


Integral 2x((1-x^2)^(1/2))

Homework Equations

The Attempt at a Solution


ive tried integration by parts, i tried substitution and i can't get it to work. i also need to figure the integral when x=1 and i keep getting 0 in the numerator. What is the best point of attack?
 
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You can use the chain rule for this integral. It is straightforward.
 
Wi_N said:

Homework Statement


Integral 2x((1-x^2)^(1/2))

Homework Equations

The Attempt at a Solution


ive tried integration by parts, i tried substitution and i can't get it to work. i also need to figure the integral when x=1 and i keep getting 0 in the numerator. What is the best point of attack?
What happens when you try ##u=1-x^2##?
 
Wi_N said:

Homework Statement


Integral 2x((1-x^2)^(1/2))

Homework Equations

The Attempt at a Solution


ive tried integration by parts, i tried substitution and i can't get it to work. i also need to figure the integral when x=1 and i keep getting 0 in the numerator. What is the best point of attack?
From the way you've written the integrand, you don't have a denominator, so getting the numerator = 0 isn't that big a deal.

It's when the denominator = 0 that things blow up. :wink: :frown:


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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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