## Area Under a Parabola

October 23, 2009 by richbeveridge

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

(2/3)(-b/a)(b²/4a-b²/2a+c)=(b³/6a²) -(2bc/3a).

Again, different.

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

0=ah²+k

-k=ah²

Plugging back into the result I had from the first go-round

A=(2/3)ah³+2hk

A=(2/3)(-k)(h)+2hk

A=(4/3)(hk)

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.

So,

A=(b³/6a²) -(bc/a)

A=(b³/6a²) -(b*0/a)

A=(b³/6a²) -0

A=(b³/6a²)

Using the formula, we got

A=(b³/6a²) -(2bc/3a), but with c=0

A=(b³/6a²) -(2b*0/3a)

A=(b³/6a²) -0

A=(b³/6a²)

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!

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on November 6, 2009 at 12:07 AM |Carnival of Mathematics #59 « The Number Warrior[…] Rich Beveridge puzzles over (sans calculus) the area under a parabola. […]

on November 6, 2009 at 11:47 AM |Pat BallewNow imagine Archimedes doing it without Newton’s help.. all with centers of gravity and such…the guy really rocked….

Nice blog,

on February 6, 2010 at 1:01 PM |thekritikorgHow would you go about deriving the formula for the parabola given its area?

on February 8, 2010 at 12:47 PM |richbeveridgeIf you choose to put one endpoint of the horizontal line at the origin, then pick where the other endpoint will be.

For instance I decided I wanted a parabola with an area of 10 and placed the other endpoint at (5, 0).

This means I need a height of 3 to produce an area of 10. So we need a y-value of three from the vertex of the parabola and a width of 5 meaning an x-value of 2.5 at the vertex.

So, we say 3=b(2.5)-a(2.5)^2

and 0=b(5)-a(5)^2

Solving for a and b we get a=-0.48 and b=2.4

So, a parabola with an area of 10 between the vertex and the x-axis would be y=2.4x-0.48x^2