Periodic function and substitution question

Click For Summary
SUMMARY

The discussion centers on proving that for a continuous periodic function f: R -> R with period T, the integral from a to a+T equals the integral from 0 to T. The key substitution involves breaking the integral into two parts: ∫ from a to a+T can be expressed as ∫ from a to T plus ∫ from T to T+a. By leveraging the periodicity of the function, specifically that f(x+T) = f(x), the equality ∫ from a to a+T f(x) dx = ∫ from 0 to T f(x) dx is established.

PREREQUISITES
  • Understanding of periodic functions and their properties
  • Knowledge of definite integrals and their evaluation
  • Familiarity with substitution techniques in calculus
  • Basic concepts of continuity in mathematical functions
NEXT STEPS
  • Study the properties of periodic functions in depth
  • Learn advanced techniques for evaluating definite integrals
  • Explore substitution methods in calculus, focusing on periodic functions
  • Review the continuity of functions and its implications on integrals
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in understanding the properties and applications of periodic functions and integrals.

cjl28
Messages
1
Reaction score
0
Let f: R-> R be a continuous function. Let T>0 be such that
f(x+T)= f(x) for all x.
We say that f is a periodic function with a period T>0.
Use an appropriate substitution to prove that for all real numbers a
[tex]\int^{a+T}_{a}f(x)dx[/tex] = [tex]\int^{T}_{0}f(x)dx[/tex].

I have no idea how to do this question.
thanks for helping me!
 
Physics news on Phys.org
cjl28 said:
Let f: R-> R be a continuous function. Let T>0 be such that
f(x+T)= f(x) for all x.
We say that f is a periodic function with a period T>0.
Use an appropriate substitution to prove that for all real numbers a
[tex]\int^{a+T}_{a}f(x)dx[/tex] = [tex]\int^{T}_{0}f(x)dx[/tex].

I have no idea how to do this question.
thanks for helping me!

Try breaking the integral on the left into two parts:

[tex]\int_{a}^{a+T} = \int_{a}^{T} + \int_{T}^{T+a}[/tex]

and in the second part, use the fact that f is periodic.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K