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### Find a Function, Integrals

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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

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,

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
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

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!  http://mathforum.org/dr.math/
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Associated Topics:
College Calculus
College Trigonometry
High School Calculus
High School Functions
High School Trigonometry

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