Question

The expression \(\frac{\int_{0}^{n}[x]dx}{\int_{0}^{n}\left \{ x\right \}dx}\), where [x] and {x} are respectively integral and fractional part of x and n∈N, is equal to

• A. \(\frac{1}{n-1}\)
• B. \(\frac{1}{n}\)
• C. n
• D. n-1

Answer 1

The Correct Option is D

The correct answer is option (D): n-1
We have, \(\int_{0}^{n}[x]dx\)
\(=\int_{0}^{}1^{0dx}+\int_{1}^{2}1 dx+\int_{2}^{3}2dx+.....+\int_{n-1}^{n}(n-1)dx\)
\(=0+1(2-1)+2(3-2)+....+(n-1)(n-(n-1))\)
1+2+3+…….+(n-1)=\(\frac{n(n-1)}{2}\)(i)
and \(\int_{0}^{n}\left \{ x \right \}dx=\int_{0}^{n}(x-[x])dx=\int_{0}^{n}xdx-\int_{0}^{n}[x]dx\)
=\(\frac{n^2}{2}=\frac{n(n-1)}{2}\)[using equation (i)]
=\(\frac{n}{2}(n-n+1)=\frac{n}{2}\)(ii)
\(\therefore \frac{\int_{0}^{n}[x]dx}{\int_{0}^{n}\left \{ x \right \}dx}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}}=(n-1)\)