Better Students Ask More Questions.
What is the definite integral of y = square root(1-x^2) if x=0 to x=1.
1 Answer | add yours
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
Posted by giorgiana1976 on April 27, 2011 at 12:12 AM (Answer #1)
Related QuestionsSee all »
Join to answer this question
Join a community of thousands of dedicated teachers and students.