Let I_{1}=int_{0}^{n}[x] d x and I_{2}=int_{0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.N

Vedclass · Accepted Answer

$n-1$

Vedclass · Answer

(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.Now, $I_{2} = \int_{0}^{n} \{x\} dx$. Since $\{x\} = x - [x]$, we have $I_{2} = \int_{0}^{n} x dx - \int_{0}^{n} [x] dx$.$I_{2} = \left[ \frac{x^2}{2} \right]_{0}^{n} - I_{1} = \frac{n^2}{2} - \frac{n(n-1)}{2} = \frac{n^2 - n^2 + n}{2} = \frac{n}{2}$.Therefore, $\frac{I_{1}}{I_{2}} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$ and $\{x\}$ are the integral and fractional parts of $x$ respectively, and $n \in N-\{1\}.$ Then, $I_{1} / I_{2}$ is equal to

Similar Questions

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$.

What is the value of $\int \limits_0^1 \cos (\pi x) \cos ([2 x] \pi) d x$? (Here $[t]$ denotes the greatest integer function of the real number $t$.)

$\int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3} = $

The value of $\int_1^2 {\log x\,dx} $ is

$\int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module