I was giving a test in my Beginning Algebra classes this past week, and while I was proctoring, I began to page through John Coburn’s Algebra and Trigonometry textbook. I came across an interesting picture indicating a formula to find the area under a parabola.
So, the formula indicates that to find the area under a parabola when it is cut by a horizontal line, we simply multiply two-thirds by the product of the length of the line segment between the points of intersection and the distance from the horizontal line to the vertex.
My first instinct was to see if I could derive this formula myself.
So, I started with a general parabola and, for the sake of simplicity, I put the left intersection point at the origin and made the horizontal line the x-axis. These choices don’t affect the generality of the derivation.
The first thing I did was to integrate the generalized parabola
∫(a(x-h)²+k)dx from 0 to 2h
This results in an answer of (2/3)ah³+2hk.
If I use the formula (2/3)(2h)(k) I get (4/3)hk. I was confused and couldn’t see right away why they were different so I integrated the parabola as ax²+bx+c instead to see what would happen.
Using the assumptions, if we integrate
∫(ax²+bx+c)dx from 0 to 2(-b/2a)=-b/a
We get (b³/6a²) -(bc/a).
If we use the formula using this method, we get
So, I looked at the assumption that the parabola goes through the origin.
If this is true in the first case (where y=a(x-h)²+k), then
Plugging back into the result I had from the first go-round
Which is what I got using the formula. OK!
Next, the ax²+bx+c method. For a parabola with the point of intersection at the origin, c=0.
Using the formula, we got
A=(b³/6a²) -(2bc/3a), but with c=0
So, again they reconcile nicely. I was glad I realized this quickly otherwise it would have stuck with me for the rest of the day!