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Finding a Function to Split the Area between Curves

Date: 01/28/2006 at 11:40:59
From: Mario
Subject: Find the area between two curves

What value of m would cause the function g(x) = mx to divide the
region bounded by f(x) = x - x^2 and the x-axis in the first quadrant
into two regions of equal area? 

So far, I found the total area bounded by y = x - x^2 and the x-axis:

A = INT[(x-x^2-x)dx] from 0 to +1

  = (x^2)/2 - (x^3)/3 - (x^2)/2 I evaluated this at 0 and +1
  = -1/3

But now I don't know what else to do.  And I think I am doing
something wrong because the area came out negative.  Can you please
give me a push?  Thank you.

Date: 01/28/2006 at 23:07:24
From: Doctor Ricky
Subject: Re: Find the area between two curves

Thanks for writing Dr. Math!

This question is a good one in that it requires us to know what an 
integral finds: the area between a curve and an axis or the area 
between two or more curves.

In this problem, we are looking for the slope of our line that cuts 
the area in half (so the area on each side of the line is equal).

We know that both of the curves begin at the origin, so they 
intersect at (0,0).  Now we need to find the other point that they 
will intersect at.

So we are looking for when:

  mx  =  x - x^2

By getting everything on the left side, we get:

  x^2 + mx - x = 0


  x^2 + (m - 1)x = 0

Now let's factor our equation:

  x*[x + (m - 1)] = 0

Since the product of these two factors equal zero, then one factor 
or the other must equal zero.

Therefore, either

  x = 0    


  x + m - 1 = 0

Since we already know that they intersect when x = 0 (the point at 
the origin), we just want to find our other x.  Solving for x, we 

  x = 1 - m      (Note this is our b value in terms of m.)

From our graph, we would say that the difference between our two 
curves (using the interval [0, 1 - m]) is equal to half the total 
area under the upper curve.  So let's put this information in 
equation form.

This means that:

     1-m                                       1
  INT   (Upper Curve - Lower Curve) = (1/2)*INT  (Upper Curve) 
     0                                         0


     1-m                              1
  INT  ( x - x^2 - mx )dx  =  (1/2)*INT  (x - x^2)dx
      0                               0

The integral on the right side is:

  (1/2)*[(x^2)/2 - (x^3)/3]

and substituting our limit values we get:

  (1/2)[(1/2) - (1/3)] = (1/2)(1/6) = 1/12   

on the right side.

Now, simplifying the left side of our equation, we get:

  (1 - m)x - x^2

Integrating, we get:

  [(1 - m)x^2]/2 - (x^3)/3.

Substituting our limit values, we get:

  (1 - m)[(1 - m)^2]/2 - [(1 - m)^3]/3

which we can see is the same as:

  [(1 - m)^3]/2 - [(1 - m)^3]/3    (look at the first term above)

This, we are saying, is equal to 1/12.

So, in our equation,

  [(1 - m)^3]/2 - [(1 - m)^3]/3 = 1/12

Multiplying both sides by 6 to get rid of our denominators, we get:

  3[(1 - m)^3] - 2[(1 - m)^3] = 1/2

which, when we combine like terms, becomes:

  (1 - m)^3 = 1/2

Taking the cube root of both sides, we get:

  1 - m = cubert(1/2)

Solving for m, we get:

  m = 1 - cubert(1/2)      


  m = 1 - 1/cubert(2)

Now let's check our answer.

What is the total area under the upper curve of f(x) = x - x^2 on 
our interval [0, 1]?

Well, we already showed it was 1/6.

What is the area between the curves f(x) and g(x) = [1 - cubert(1/2)]
x on the interval [0, cubert(1/2))]?  (Read back when we found our x 
value why this is our b value!)  

After we evaluate this integral, we get that this area equals 1/12.

Since our total area is 1/6 and this area is 1/12, we have found the 
m that results in cutting our area perfectly in half!

If you have any questions or if this was a little confusing, please 
let me know!

- Doctor Ricky, The Math Forum 

Date: 01/29/2006 at 12:30:56
From: Mario
Subject: Thank you (Find the area between two curves)

Thank you so much Doctor Ricky, you have no idea how much this helped
me.  I understand everything you did and it looks simple, now that you
explained it to me.  I have other problems similar to this one, and I
am solving them with little or no difficulty.  Thank you!
Associated Topics:
College Calculus

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