In order to simplify the problem, suppose we have a parabola $y=ax^2+bx+c$, here $a\neq0$, and a line $y=kx+d$, here $k\neq0$. We can assume that they will intersect at two different points. Thus, the $\Delta$ of the equation $ax^2+bx+c=kx+d$ will be greater than $0$($\Delta> 0$). Let $S$ be the area closed by them, it is clear that $S>0$.
Now I wonder how to calculate $S$ without calculus?
graph http://i.minus.com/ibzmus78w1n2Qr.png
UPDATE:
I try to solve this problem without calculus in order to make my little brother who don't know about calculus understand it. You can solve it as long as you can make a junior student understand your solution. 🙂
Best Answer
Archimedes derived a formula for this area. (It doesn't use calculus, which wouldn't be invented until centuries later.)
The area enclosed by a parabola (with vertical axis) and a chord AB is 4/3 the area of the triangle ABC where C is the point on the parabola whose x-coordinate is halfway between the x-coordinates of A and B. This result and Archimedes' method of deriving it are in the Wikipedia article The Quadrature of the Parabola.
I don't know if this will be any easier for your little brother to understand than a solution with calculus, but let us know.