Wednesday, February 6, 2013

What is the definite integral of y = square root(1-x^2) if x=0 to x=1.

I'll suggest to replace the variable x by the function sin
t.


x = sin t


We'll
differentiate both sides and we'll get:


dx = cos t
dt


We'll change the limits of integration,
too:


x = 0 => t = 0


x =
1 => t = pi/2


The definite integral will be
evaluated using the Leibniz Newton formula:


Int sqrt[1 -
(sin t)^2] cos t dt = F(pi/2) - F(0)


But, from Pythagorean
identity, we'll have:


1 - (sin t)^2 = (cos
t)^2


We'll take square root both
sides:


sqrt [1 - (sin t)^2] = sqrt (cos
t)^2


sqrt [1 - (sin t)^2] = cos
t


Int sqrt[1 - (sin t)^2] cos t dt = Int (cos t)^2
dt


Int (cos t)^2 = Int (1+cos
2t)dt/2


Int (1+cos 2t)dt/2 = Int dt/2 + (1/2)*Int cos 2t
dt


Int (1+cos 2t)dt/2 = t/2 + sin
2t/4


F(pi/2) - F(0) = pi/4 + sin pi/4 - 0 + sin
0/4


sin pi = 0


F(pi/2) - F(0)
= pi/4


The definite integral of the given
function is: Int sqrt(1 - x^2) dx = pi/4

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