Question

If $f(x)$ is a continuous periodic function with period T, then the integral $I = \int_{a}^{a + T}{f(x)dx}$ is

• A. Equal to $2a$
• B. Equal to 3a
• C. Independent of a
• D. None of these

Answer 1

Answer: (C)

Independent of a

Answer 2

Consider the function $g(a) = \int_{a}^{a + T}{f(x)dx}$ =

$\int_{a}^{0}{f(x)dx} + \int_{0}^{T}{f(x)dx} + \int_{T}^{a + T}{f(x)dx}$

Putting $x - T = y$ in last integral, we get

$\int_{T}^{a + T}{f(x)dx} = \int_{0}^{a}{f(y + T)dy} = \int_{0}^{a}{f(y)dy}$

⇒ $g(a) = \int_{a}^{0}{f(x)dx} + \int_{0}^{T}{f(x)dx} + \int_{0}^{a}{f(x)dx}$ = $\int_{0}^{T}{f(x)dx}$

Hence $g(a)$ is independent of a.