Associated Topics || Dr. Math Home || Search Dr. Math

### Area Between a Line and a Curve

```
Date: Sat, 5 Nov 1994 17:13:46 -0500
Subject: math problem

Here is a problem that has been bugging me:
"The horizontal line y=c intersects the curve y=2x-3x^3 in the first
quadrant.  Area A is bounded by y=c, x=0, and the curve.  Area B is
bounded above by the curve and below by y=c.  Find c such that
(area of A)=(area of B)."

I can do this using integrals, but I don't want to solve any @!%\$! cubics.
There must be an easier way to find c.

AP
ps I am a senior at Athens High School.
```

```
From: Dr. Ken
Date: Mon, 7 Nov 1994 17:33:47 -0500 (EST)

AP!

Nice problem!  At first glance, it seems so easy, and then you think it's
virtually unsolvable.  Try this, though:

You need to get a handle on the limits of integration, i.e. what values of x
you can plug in to f(x) to get c.  I'll call these c1 and c2.  Then they
satisfy the equation 2c1 - 3c1^3 = 2c2 - 3c2^3.

So 2(c1 - c2) - 3 (c1^3 - c2^3) = 0, and since x^3 - y^3 = (x-y)(x^2+xy+y^2),
we can factor out a (c1 - c2) from this equation (it yields only the trivial
root).  Then you get a quadratic equation relating c1 and c2, and you can
solve for c1 in terms of c2.  It's nasty, but it works.

I haven't taken this any farther, but I think this is going to help.  If it
doesn't, write back and we'll see what we can do.

-Ken "Dr." Math
```
Associated Topics:
High School Calculus

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search