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Mathematics Question on Some Properties of Definite Integrals

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

1n−1\frac{1}{n-1}

B

1n\frac{1}{n}

C

n

D

n-1

Answer

n-1

Explanation

Solution

The correct answer is option (D): n-1
We have, ∫0n[x]dx\int_{0}^{n}[x]dx
=∫010dx+∫121dx+∫232dx+.....+∫n−1n(n−1)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))=0+1(2-1)+2(3-2)+....+(n-1)(n-(n-1))
1+2+3+…….+(n-1)=n(n−1)2\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
=n22=n(n−1)2\frac{n^2}{2}=\frac{n(n-1)}{2}[using equation (i)]
=n2(n−n+1)=n2\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)