For example, this integral is equal to Pi:
http://i.imgur.com/tJS1ZaA.png
It's been years since I've done any calculus. Do I understand that integral correctly as finding the area of a circle of radius 1, which is π?
http://i.imgur.com/tJS1ZaA.png
It's been years since I've done any calculus. Do I understand that integral correctly as finding the area of a circle of radius 1, which is π?
f(x) = sqrt(1-x^2)
2 * integrate(f(x),x,-1,1)
plot(f(x),x,-1,1,gridlines=True,aspect_ratio=1)
The function by itself produces a line equal to 1/2 the distance across a unit circle at the coordinate provided by x. Therefore a definite integral of the function on the interval -1 to 1 should produce an area equal to 1/2 that of a unit circle, or pi/2.
My Calculus tutorial: http://www.arachnoid.com/calculus/
http://www.wolframalpha.com/input/?i=2+integrate%28sqrt%281-...
The formula for a circle's area is a = PI r^2. For a unit circle, the area is simply equal to PI.
Now for the right side of the equation:
r^2 = x^2 + y^2 is the Pythagorean Theorem and also the equation of a circle. If we solve for y, we have y = sqrt(r^2 - x^2). In other words, the height of any circle (centered at the origin) is sqrt(r^2 - x^2) given any particular place along the x-axis. This is the same expression which is integrated.
For a unit circle, we assume r = 1 and ignore everything outside of -1 and 1 on the x-axis. This is why the integral is bound between -1 and 1. So now we have integrate(-1, 1){ sqrt(1^2 - x^2) }.
By integrating we get the area of the top half of the unit circle. But we also need the bottom half. So we just double everything and that's what the extra 2 is for. We now have the area of a unit circle on the right side too.
QED