Studying Mathematics Thread

[GrandizerII]

Remember that the 'area function' is an anti-derivative.

So, the definite integral from 0 to x is an area under the circle. Draw a radius to the point on the circle, which divides the area into two pieces.

The lower piece is a triangle. It's area is (1/2) xy=(1/2) x sqrt(1-x^2)

The upper piece is an angular sector. The area is (1/2) rA, where A is the radian measure of the angle in question. But, r=1, and the picture shows that sin(A)=x. Hence,
A=sin^{-1}(x).

Putting the areas together gives an anti-derivative of sqrt(1-x^2) of

(1/2)x sqrt(1-x^2) + (1/2) sin^{-1} (x)

For the general anti-derivative, add +C.

Interesting. Not sure I would've figured this out eventually on my own, but this is refreshing to have a new look on this, instead of always being exposed to the standard solutions available on YouTube and such.