The expression frac{int_0^n [x] dx}{int_0^n { | vedclass

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(D) Let $I_1 = \int_0^n [x] dx$ and $I_2 = \int_0^n \{x\} dx$. $I_1 = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \dots + \int_{n-1}^n (n-1) dx$.

Vedclass · Accepted Answer

$n-1$

Vedclass · Answer

(D) Let $I_1 = \int_0^n [x] dx$ and $I_2 = \int_0^n \{x\} dx$. $I_1 = \int_0^1 0 dx + \int_1^2 1 dx + \int_2^3 2 dx + \dots + \int_{n-1}^n (n-1) dx$. $I_1 = 0 + 1 + 2 + \dots + (n-1) = \frac{(n-1)n}{2}$. Since the function $\{x\}$ is periodic with period $1$, we have $I_2 = n \int_0^1 \{x\} dx = n \int_0^1 x dx$. $I_2 = n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$. Therefore, the expression is $\frac{I_1}{I_2} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

The expression $\frac{\int_0^n [x] dx}{\int_0^n \{x\} dx}$, where $[x]$ and $\{x\}$ are respectively the integral and fractional part of $x$ and $n \in N$, is equal to

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