Definite Integration Question 255
Question: The value of the integral $ \sum\limits _{k=1}^{n}{\int_0^{1}{f(k-1+x)dx}} $ is
Options:
A) $ \int_0^{1}{f(x)dx} $
B) $ \int_0^{2}{f(x)dx} $
C) $ \int_0^{n}{f(x)dx} $
D) $ n\int_0^{1}{f(x)dx} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ I=\int_0^{1}{f(k-1+x)dx} $
therefore $ I=\int _{k-1}^{k}{f(t)dt,} $ where $ t=k-1+x $
therefore $ I=\int _{k-1}^{k}{f(x)dx} $
$ \therefore \sum\limits _{k=1}^{n}{\int _{k-1}^{k}{f(x)dx=\int_0^{1}{f(x)dx+\int_1^{2}{f(x)dx+…..+\int _{n-1}^{n}{f(x)dx}}}}} $
$ =\int_0^{n}{f(x)dx} $ .
BETA
