Evaluate $\int_0^3\sqrt{9-x^2}dx$ using areas
If I divide $[0,3]$ into $n$ equal partitions, and chose the height of the partition as the right end point, then
$$\int_0^3\sqrt{9-x^2}dx=\sum_{k=0}^n\sqrt{9-k^2}\cdot \frac{3}{n}$$
But the riemann sum is difficult to evaluate. How should I choose my partition to make the sum easier to solve?
Actually $f(x) = \sqrt{9-x^2}$ for $x \in [0,3]$ represents a circle of radius $3$ restricted in the first quadrant. This is because the implicit equation of the circle is $x^2 + y^2 = 3^2 \implies y = \pm \sqrt{9-x^2}$. Therefore the integral is simply the area of the quarter of the circle : $9\pi/4$.