I found a fascinating page: VisualCalc. This is a talk by Tom Apostol about Visual Calculus, a technique for finding the area bounded by curves without using traditional calculus developed by an Armenian mathematician living in California, Mamikon A. Mnatsakanian, which has been espoused by Apostol. Some of the results using this method would be very difficult if not impossible to uncover with traditional methods. The starting point is the following simple (but neat) problem, solved by Mamikon (as he calls himself) when he was 15:
Problem: A line segment is drawn tangent to the inner of two concentric circles, terminating at the outer circle. The length of the segment is 2a. What is the area of the annulus?
Answer: pi*a^2. It is rather counterintuitive that the result is independent of the radius of the inner circle.
Solution: Let the radius of the smaller and larger circles by r and R, respectively. The area of the annulus is pi*(R^2 - r^2). Draw the obvious right triangle with legs of length r and a, and hypotenuse of length R. Apply the Pythagorean Theorem.
Mamikon noted that if he knew in advance that the answer was independent of r, he could let r = 0, and the tangent segment would become a diameter of the larger circle, establishing the result another way. This led him to a rather breathtaking extension of the result.
Theorem 1. Let C be a smooth convex oval. Move a vector v (of fixed length) around the oval (with the tail on the curve) so that it is always tangent to the curve (at its tail). Then the area swept out by the vector is pi*|v|^2. [I'm not sure what the exact hypothesis is, but this is the basic idea.]
Proof idea: Let S be the set of translates of the vectors v(t), with a common tail formed as v(t) goes around the oval. Then S is a circle of radius |v|.