Question

How do you find the definite integral for: `(sqrt(b^2-a^2)) da` for the intervals `[0, b]`?

Answer 1

Used the geometric meaning of the definite integral.

Explanation:

The graph of `y = sqrt(r^2-x^2)` is the upper semicircle centered at `(0,0)` with radius `r`

The integral from `0` to `r` is, therefore one fourth of the area of the circle with radius `r`.

In this problem, we have changed the independent variable to `a` and the name of the radius to `b`, but the reasoning is unchanged.

`int_0^b sqrt(b^2-a^2) da = 1/4(pi b^2)`