Definite Integration Question 476
Question: If $ f(x) $ is a continuous periodic function with period $ T, $ then the integral $ I=\int_a^{a+T}{f(x)dx} $ is
Options:
A) Equal to $ 2a $
B) Equal to $ 3a $
C) Independent of $ a $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
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}}} $
therefore $ g(a)=\int_a^{0}{f(x)dx+\int_0^{1}{f(x)dx+\int_0^{a}{f(x)dx}}} $
$ =\int_0^{T}{f(x)dx} $
Hence g is independent of a.
BETA
