The 'integration by change of variable' theorem

In summary, the 'integration by change of variable' theorem states that if g is a continuous function on [c,d], and g' is continuous on [c,d] but g([c,d]) is not a subset of [a,b], then [tex]\int_a^b f(x)dx = \int_c^d f(g(t))g'(t)dt[/tex]
  • #1
quasar987
Science Advisor
Homework Helper
Gold Member
4,807
32
In my analysis textbook, the 'integration by change of variable' theorem reads

"Theorem: Consider f a continuous fonction on [a,b], g a continuous function on [c,d] such that g' is continuous on [c,d]. If g([c,d]) is a subset of [a,b] and if g(c) = a and g(d) = b, then

[tex]\int_a^b f(x)dx = \int_c^d f(g(t))g'(t)dt[/tex]


But reading the proof, I nowhere see the need for g([c,d]) being a subset of [a,b]. Is this really necessary? As long as g(c) = a and g(d) = b, it should be alright, no?
 
Physics news on Phys.org
  • #2
No. Suppose a is 0, b= 1, and g(x) is the function g(x)= 8x(1-x)= 8x- 8x2. To make it simple, suppose f is the constant function f(x)= 1.

Then g'(x)= 8- 16x and the theorem, without that requirement, would assert that
[tex]\int_0^1dx= 1= \int_0^1(8- 16x)dx= 0!
 
  • #3
But you use [0,1] has your interval [c,d]. And while g(0) = 0, g(1) does not equal 1. It's 0 too.
 
Last edited:
  • #4
*bump*

Yes, I'm still curious about this also. It looks like the definition of a line integral in complex analysis:

If

[tex]\int_{\gamma}f(z)dz=\int_a^bf(g(t))g'(t)dt[/tex]

...g(t) is the function for the parametrization of the curve.

right? And in this case, we know that the parametrization does not matter: as long as you start and end at the correct points.

(Also, forgive my LaTeX!)
 
  • #6
It makes sense that it shouldn't matter. Think of a very simple case where g(x) on [c,d] has a single maximal point at c<m<d and say g(c)<g(d)<g(m). So there is a c1, c< c1 < m such that g(c1)=g(d). So since this is a simple case we'll say g([c,c1]) is a subset of [g(c),g(d)]. So
[tex]\int_{g(c)}^{g(d)} f(x) dx = \int_c^{c1} g'(x)g(x)dx[/tex] and
[tex]\int_{c1}^m g'(x)g(x)dx = \int_{g(c1)}^{g(m)} f(x) dx = -\int_{g(m)}^{g(c1)} f(x) dx=- \int_{m}^d g'(x)g(x)dx[/tex]
So the sum cancels out nicely.

Remember this is a simplified case and in no way am I saying this is a proof for the general case. You can have infinite oscillations so that you are dealing with an infinite sum with no guarantee (from me at least) that it will converge where you want it to.
 

Related to The 'integration by change of variable' theorem

What is the "integration by change of variable" theorem?

The "integration by change of variable" theorem, also known as the substitution rule, is a mathematical technique used to evaluate integrals by replacing the variable of integration with a new variable.

Why is the "integration by change of variable" theorem useful?

The "integration by change of variable" theorem is useful because it simplifies the process of evaluating integrals that would otherwise be difficult or impossible to solve using traditional methods.

What is the process for using the "integration by change of variable" theorem?

The process for using the "integration by change of variable" theorem involves identifying a suitable substitution, making the substitution, and then evaluating the resulting integral in terms of the new variable.

What are some common substitutions used in the "integration by change of variable" theorem?

Some common substitutions used in the "integration by change of variable" theorem include trigonometric substitutions, u-substitutions, and exponential substitutions.

Can the "integration by change of variable" theorem be used to solve all types of integrals?

No, the "integration by change of variable" theorem is not applicable to all types of integrals. It is most useful for integrals involving algebraic, trigonometric, or exponential functions.

Similar threads

  • Calculus
Replies
6
Views
1K
Replies
4
Views
849
  • Calculus
Replies
9
Views
2K
Replies
1
Views
1K
Replies
7
Views
1K
  • Calculus
Replies
29
Views
871
  • Calculus and Beyond Homework Help
Replies
2
Views
269
Replies
2
Views
390
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top