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Saturday, August 21, 2010

Calculus Forgotten

Today over lunch my friend and I challenged each other with math questions. Below is the one he gave me. To my embarassment, I had forgotten how to even begin to solve it...

Integral from 0 to 1 of the square root of 1-x^2

Well, do you remember?

3 Comments:

  • i'm kinda rusty, but i think you need to solve that via trig substitution.

    ∫ (1-x^2)^1/2 dx

    assume x = sin k; then dx = cos k dk ->

    ∫ (1-(sin k)^2)^1/2 (cos k dk) ->
    ∫ (1-sin^2 k)^1/2 (cos k dk) ->
    ∫ (cos^2 k)^1/2 (cos k dk) ->
    ∫ cos k (cos k dk) ->
    ∫ cos^2 k dk
    ...

    i think you should end up with 1/2(x(1-x^2)^(1/2) + sin^-1 x) + c

    By Blogger w, at 8/21/2010 10:35 PM  

  • The primitive is:

    F(x) = 1/2 * (sqrt(1-x^2)*x + arcsin(x))

    So F(x)|(0,1) = F(1) - F(0) = pi / 4

    By Anonymous Alex Ksikes, at 10/27/2010 2:10 AM  

  • @w, @Alex - Well guys, your calculus is better than mine, I must say.

    By Blogger Jonathan, at 7/21/2011 8:09 PM  

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