Expert: Paul Klarreich - 2/28/2007

Question
Hi Mr. Klarreich!

Im in Calculus 2 and I'm trying to solve a bunch of these word problems in my calculus class and cant figure out how to find the area.

1) Find the number "a" such that the line "x=a" bisects the area under the curve:

y=1/x^2 , 1 is less than or equal to x which is less than or equal to 4.

2) Find the number b such taht the line "y=b" bisects the area in the previous problem.

SO far this is what I know --first, find the area under the curve 1/x2 from x = 1 to x = 4 setting this as an integral but I dont know how...Then I know that you can, from the area cut it in half as A/2

At some point I also know to use the fundamental theorem of calculus

As for the second part, I can use the area from A and then solve for b.

Thank you a lot for your help on this problem!

Answer
Questioner:   Alison
Category:  Calculus
 
Subject:  Calculus and area under curve
Question:  Hi Mr. Klarreich!

Im in Calculus 2 and I'm trying to solve a bunch of these word problems in my calculus class and can't figure out how to find the area.

1) Find the number "a" such that the line "x=a" bisects the area under the curve:

y=1/x^2 , 1 <= x <= 4.

2) Find the number b such that the line "y=b" bisects the area in the previous problem.

SO far this is what I know --first, find the area under the curve 1/x2 from x = 1 to x = 4 setting this as an integral but I dont know how...Then I know that you can, from the area cut it in half as A/2

At some point I also know to use the fundamental theorem of calculus As for the second part, I can use the area from A and then solve for b.

Thank you a lot for your help on this problem!
........................................................
Hi, Ali,

[Sorry, but that's what we all call our niece with the same name.]
WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

For your #1,

The area from x = 1 to x = a is given by:

{a  1
|  --- dx =
}1 x^2

{a  
|  x^-2 dx =
}1
x^-1
---- from 1 to a =
-1

-1
--- from 1 to a =
x

(-1/a - (-1/1)) = -1/a + 1

The area from x = a to x = 4 is given by:

{4  1
|  --- dx =
}a x^2

{4  
|  x^-2 dx =
}a
x^-1
---- from a to 4 =
-1

-1
--- from a to 4 =
x

(-1/4 - (-1/a)) = -1/4 + 1/a

Now those two are supposed to be equal.
-1        -1     1
--- + 1 = --- + ---
a         4     a
Solve for a:
5     2
--- = ---
4     a
5a = 8
a = 8/5

....................................
For the second one, you want a horizontal bisection.  This looks trickier, so let's consider it this way:

A. The entire area is given by:

{4  1
|  --- dx =
}1 x^2

{4  
|  x^-2 dx =
}1
x^-1
---- from 1 to 4 =
-1

-1
--- from 1 to 4 =
x

(-1/4 - (-1/1)) = -1/4 + 1

= 3/4

B. Now what value of  y = b will produce a bisection?  This is really an 'area between two curves' problem; the upper curve is y = 1/x^2, and the lower is y = b.   

The graph of y = 1/x^2 is descending and we can expect the horizontal line to cross it at the right.  Where will it cross?

Suppose we integrate from x = 1 to x = r.  When x = r, y = 1/r^2, which is our  b.

{r
|  (x^-2 - 1/r^2) dx
}1
 -1     x
= --- - ---, from 1 to r.
  x    r^2

  1     x
= --- + ---, from r to 1. << SWITCHING SIDES TRICK.
  x    r^2

(1 + 1/r^2) - (1/r + 1/r) = 3/8

1 + 1/r^2 - 2/r = 3/8

5/8 + 1/r^2 - 2/r = 0

5r^2 - 16r + 8 = 0

This is a quadratic equation, and there is nothing to be done for it except:

   16 +- sqrt(256 - 160)
r = ---------------------
           10

   16 +- sqrt(96)
r = ---------------
         10

   16 +- 4 sqrt(6)
r = ---------------
         10

   8 +- 2 sqrt(6)
r = ---------------
         5

We need the positive sign there.  If you try using the negative sign you will get a root that is <1; no good, because it is outside the [1,4] interval of the problem.

   8 + 2 sqrt(6)
r = ------------- ~~ 2.5797959
        5
And our value of b:

b = 1/r^2
         25
= --------------------
 64 + 24 + 32 sqrt(6)
 
       25
= ---------------- ~~ 0.150255
 88 + 32 sqrt(6)

Sorry this didn't come out neater, but I have done some numerical checking and these results seem to work out.