Question

Which of the following statements are true?

• A. If f(x) be continuous and periodic with periodicity T, Then $I=\int_{a}^{a+T}f(x)dx$ depend on 'a'.
• B. If f(x) be continuous and periodic with periodicity T, Then $I=\int_{a}^{a+T}f(x)dx$ does not depend on 'a'.
• C. Let f(x)=$\begin{Bmatrix} 1 &if \,x\, \,is \,rational \\\ 0&if \,x\, \,is \,irrational \end{Bmatrix}$, then f is periodic of the periodicity T only if T is rational
• D. f defined in (C) is periodic for all T.

Answer 1

Answer: (B), (C)

If f(x) be continuous and periodic with periodicity T, Then $I=\int_{a}^{a+T}f(x)dx$ does not depend on 'a'.

Answer 2

A function f(x) is said to be periodic with periodicity T if f(x +T)=f(x) for all x. This means that the function has the same values at intervals of T.

When you consider the integral =∫=∫ aa +T ​ f(x)dx , you are integrating f(x) over a complete period of the function. Since f(x) is periodic, integrating over one period will yield the same result regardless of the starting point a. This is because the values of f(x) repeat every T , so the integral will accumulate the same area under the curve regardless of where you start within a period.

Mathematically, this can be expressed as: ∫=∫0∫ aa +T ​ f(x)dx =∫0 T ​ f(x)dx where 00 to T represents one complete period of the function.

In essence, since f(x) repeats itself every T , integrating over any interval of length T will give you the same result. Therefore, the integral I does not depend on the choice of a.

The correct answer is/are option:
(B) : If f(x) be continuous and periodic with periodicity T, Then $I=\int_{a}^{a+T}f(x)dx$ does not depend on 'a'.
(C): Let f(x)=$\begin{Bmatrix} 1 &if \,x\, \,is \,rational \\\ 0&if \,x\, \,is \,irrational \end{Bmatrix}$, then f is periodic of the periodicity T only if T is rational