Definite Integration Question 446
Question: If n is a positive integer and [x] is the greatest integer not exceeding x, then $ \int_0^{n}{{x-[x]}dx} $ equals
Options:
A) $ n^{2}/2 $
B) $ n(n-1)/2 $
C) $ n/2 $
D) $ \frac{n^{2}}{2}-n $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x-[x] $ is a periodic function with period 1.
$ \therefore \int_0^{n}{{ x-[x] }dx=n\int_0^{1}{(x-[x])dx}} $
$ =n[ \int_0^{1}{xdx-\int_0^{1}{[x]dx}} ] $
$ =n[ ( \frac{x^{2}}{2} )_0^{1}-0 ]=\frac{n}{2} $ .
BETA
