The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Find a Function, Integrals

Date: 12/17/97 at 18:59:32
From: Lanka
Subject: Integrals

Hi Dr. Math,

I have a couple of questions regarding integrals. The first question 
reads as follows:

Suppose the graph of f has the formula

  f(x) =  -x + 1 for 0 <= x <= 1
           x - 1  for 1 < x <= 2

  a) Find a function F such that F' = f and F(1) = 1
  b) use geometry to show that the area under the graph of f above 
     the x-axis between x = 0 and x = 2 is equal to F(2)-F(0)

What I did:

I found a function F such that F' = f and F(1) = 1:

    F = |(x^2)/2 - x + (3/2)|
   F' = |x - 1|

However, by the graph of f the area under the curve above the x axis
from x = 0 to x = 2 should be equal to 1.  But F(2)- F(0) gives me 
zero, what have I done wrong here?

My second question is finding the integral of the following function
using the process of substitution:

   g(theta) = sin(theta) + (1/(1+ theta^2))

Date: 12/29/97 at 10:21:26
From: Doctor Mark
Subject: Re: Integrals

Hi Lanka,

Let's look at your first problem. Your answer for what F is so that 
F' = |x-1| is *a* correct one, but not the only possible answer.  

First of all, note that the derivative, f, of F is defined in a 
"piecewise" fashion, i.e., the function f has one algebraic expression 
for some values in the domain ((0 <= x <= 1) (let's call that Region 
I), and another algebraic expression for values in another domain 
(1 < x <=2) (I'll call that Region II). If we were to look for a 
function that has f = -x + 1 in Region I, then the general function F 
with that derivative would be

   F(x) = - (x^2)/2 + x + C1 (for x in Region I),

where C1 is some constant.

If the function is f = x - 1 in Region II, then the general function F
with that derivative in Region II would be

   F(x) = (x^2)/2 - x + C2 (for x in Region II),

where C2 is some constant.

So this the general "piecewise" algebraic expression for F(x). Let's 
see what C1 and C2 are. We know that F(x) is supposed to be continuous 
at x = 1, and in fact that when x = 1, the two functions must both 
give us the value F(1) = 1. Substituting x = 1 into both of the 
definitions, we find that

   C1 = 1/2

   C2 = 3/2.

So here is the definition of F(x):

              - (x^2)/2 + x + 1/2  for 0<= x <=1
   F(x)  =
                (x^2)/2 - x + 3/2   for 1 < x <=2

The mistake you made was to assume that the two constants had to be 
the same for each region, and they are not. I've made that kind of 
mistake myself, so you shouldn't feel too bad: you have to be careful 
when the derivative of a function is defined in a piecewise way like 
this (and in particular, whenever the derivative is defined through an 
absolute value, which is really a "piecewise" defined function).

Now we can check that this function gives the right results for 
F(2) - F(0):

For F(2) we have to use the *lower* function above, since that is what 
F is when 1 < x <=2  (which x=2 certainly is!), so

   F(2) = (2^2)/2 - 2 + 3/2 = 3/2

And for F(0), we have to use the *upper* function above, since that is 
what F is when 0<=x<=1, which x = 0 certainly is:

   F(0) = -(0^2)/2 + (0) + 1/2 = 1/2.

Hence, F(2) - F(0) = 3/2 - 1/2 = 1, as desired.

So this is the correct answer.  What's wrong with yours?  Well, if you 
look at the quantity inside the absolute value sign in your expression 
for F, that would be

   G = (x^2)/2 - x + 3/2.

But if you were to graph that function, you would find that it is a
parabola that never dips below the x-axis (you could show that either 
by using a graphing calculator, or algebraically by factoring out a 
1/2 and completing the square: you find that

   G(x) = [x^2 - 2X + 3]/2 = [(X-1)^2 + 2]/2,

which is manifestly positive for any value of x, because of the 
(x-1)^2 .

That is, G is never negative, and so taking the absolute value of it 
does nothing at all.  Hence, its derivative is just  x - 1, for *all* 
x, and this is not the derivative you want.

The assumption that you made was that if the derivative is given by an
absolute value, so is the function, and that is in general not true, 
as we just saw.

Having said this, you *must* understand that this is a *very* subtle 
point that could have tripped up a lot of professional mathematicians, 
so you should not feel too bad about it. As they say, you live and you 
learn (from your mistakes).

As for your second problem, whenever you see something like 
1/(1 + (something)^2), the denominator should remind you of a 
trigonometric identity of the same form, namely:

   1 + tan^2 = sec^2

(which is really just the Pythagorean Theorem, sin^2 + cos^2 = 1, 
disguised by dividing it by cos^2). If you let theta = tan x, you will 
see how to do the second integral. Try it!

(Of course, the first term of your integrand is easy to do.)

I hope this has been of some help.  If you have any other questions, 
be sure to write back.

-Doctor Mark,  The Math Forum
 Check out our web site!   
Associated Topics:
College Calculus
College Trigonometry
High School Calculus
High School Functions
High School Trigonometry

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

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.