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...
Well, do you remember?
Well, do you remember?
posted by Jonathan at
8/21/2010 08:26:00 p.m.
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
w, at 8/21/2010 10:35 p.m.
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
Alex Ksikes, at 10/27/2010 2:10 a.m.
@w, @Alex - Well guys, your calculus is better than mine, I must say.
By
Jonathan, at 7/21/2011 8:09 p.m.
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