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Why Does Integration by Substitution Work?

Date: 12/13/2006 at 22:03:56
From: Reuben
Subject: Why integration by substitution works

I just learned the process of integration by substitution.  My
textbook seems to explain it for the most part, but it says "imagine
that you multiply du/dx by the dx in the derivative to get du."  This
seems to work, but I've found online that technically, you can't treat
differentials as numbers that can be "canceled out."

A) Is this true?  It seems to me that they're just very very very
small numbers.  So why could you not cancel them out?

B) If that's the case, how do you prove that substitution works and
use integration by substitution WITHOUT treating differentials as numbers?



Date: 12/14/2006 at 08:39:02
From: Doctor Fenton
Subject: Re: Why integration by substitution works

Hi Reuben,

Thanks for writing to Dr. Math.  There are several ways to look at
differentials, but probably the simplest is to view them as a formal
bookkeeping device for keeping track of the constants when finding
antiderivatives.

Many of the integration (or antidifferentiation) rules are actually
counterparts of corresponding differentiation rules, and this is true
of the substitution theorem, which is the integral version of the
Chain Rule.

The Chain Rule says that if we have a composite function F(g(x)), and
if f(x) = F'(x) is the derivative of the outer function F(x), then

   [F(g(x))]' = F'(g(x))*g'(x)
              = f(g(x))*g'(x) .

The antiderivative version of this says that if we want to find the
antiderivative of f(g(x)) and we know the antiderivative F of f, then
the antiderivative of f(g(x)) is just F(g(x)), and we have reduced the
problem of finding the antiderivative of the complicated expression
f(g(x)) to that of finding the antiderivative of f, which we usually
write with a different independent variable such as u.  That is, the
Chain Rule says that

   [F(g(x))]' = f(g(x))*g'(x) ,  (remember that F' = f)

so

   /
   | f(g(x))*g'(x)dx = F(g(x)) + C .
   /

But since F' = f, 

  /
  | f(u)du = F(u)
  /

(the letter used for the variable of integration is a dummy variable,
so we can use any letter we wish).  This lets us write

   /                   /
   | f(g(x))*g'(x)dx = | f(u)du
   /                   /

where we understand the right side to be F(u) with u replaced by the
formula g(x).  So it appears that in the integral on the left side, we
replaced g(x) by u and g'(x)dx by du (similar to the way it "appears"
that the sun rises and sets).  This process doesn't have to be
considered meaningful in itself, but rather just a mnemonic or aid in
determining the outer function f(x) in the composition.  The
differential part of this substitution can also be thought of as a
mnemonic, since we can mix Leibnitz and function notation to write, if
u = g(x), that

   du
   -- = g'(x)   ,
   dx

and "multiplying" by dx gives  du = g'(x)dx .  One can make this 
procedure logically rigorous by introducing the concept of 
differential forms, but that requires a lot of mathematical machinery, 
so that it is easier to just think of it as a formal computation.  
(There are books such as Bressoud's _Second Year Calculus_, Edwards' 
_Advanced Calculus: A Differential Forms Approach_, and Spivak's 
_Calculus on Manifolds_ which go over this in detail.)

For example, to integrate

   /
   | [x^3+1]^(1/2) x^2 dx ,
   /

we can notice that the integrand has the form of the derivative of 
a composite function f(g(x))*g'(x), with f(x) = x^(1/2) and 
g(x) = x^3+1, but g'(x) = 3x^2, not just x^2, so the constant is not
quite right.  However, if we multiply and divide by 3, we have

   /                             /
   | [x^3+1]^(1/2) x^2 dx = (1/3)| [x^3+1]^(1/2)*(3x^2)dx
   /                             /

and the integrand on the right is exactly a derivative now, of the
composite function (2/3)[x^3+1]^(3/2) (we can see this because the
outer function f(x) is x^(1/2), whose antiderivative F(x) is
(2/3)x^(3/2), so the antiderivative is (1/3)F(g(x))).  This argument
has computed the integral or antiderivative without substitution.

To use substitution, we let u = x^3+1, and du = 3x^2 dx, so that
x^2dx = (1/3)du, and

   /                        /
   | [x^3+1]^(1/2) x^2 dx = | u^(1/2) (1/3)du
   /                        /
                            
                          = (1/3) * (2/3)u^(3/2) + C
 
                          = (2/9) [x^3+1]^(3/2) + C  .
                            
Substitution makes the process fairly mechanical so it doesn't
require much thought, once you see the appropriate substitution to
use, and it also automatically keeps the constants straight. 

The objective of indefinite integration is to find an antiderivative,
and exactly how you do that isn't really important, at least in my
opinion.  If you can just look at a formula and "see" what the 
antiderivative is, you have solved the problem.  Most of us can't do
that, and there are a number of procedures to help, such as the
substitution rule, and integration by parts (the integral version of
the Product Rule of differentiation).  The real theory behind
substitution is the Chain Rule, and you can look at the details of
substitution as a formal process for helping you see the important
parts of the composite functions involved, without worrying about
their intrinsic meaning.

If you have any questions, please write back and I will try to 
explain further.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 12/14/2006 at 13:45:51
From: Reuben
Subject: Thank you (Why integration by substitution works)

Thank you so much.  Your response has made the reasoning behind this
SO much clearer to me!
Associated Topics:
College Calculus
High School Calculus

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