Let n be a positive integer. For a real numbe | vedclass

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

Question

(C) Let $I = \int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$. Since $[x]$ is constant on intervals $[k, k+1)$ for $k \in \{1, 2, \ldots, n\}$,we c

Vedclass · Accepted Answer

$\frac{n}{n+1}$

Vedclass · Answer

(C) Let $I = \int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$. Since $[x]$ is constant on intervals $[k, k+1)$ for $k \in \{1, 2, \ldots, n\}$,we can split the integral as: $I = \sum_{k=1}^{n} \int_{k}^{k+1} \frac{(\{x\})^k}{k} d x$. Substituting $u = x-k$,so $dx = du$,and $\{x\} = u$ as $x$ ranges from $k$ to $k+1$: $I = \sum_{k=1}^{n} \frac{1}{k} \int_{0}^{1} u^k du = \sum_{k=1}^{n} \frac{1}{k} \left[ \frac{u^{k+1}}{k+1} \right]_0^1 = \sum_{k=1}^{n} \frac{1}{k(k+1)}$. Using partial fractions,$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Thus,$I = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$. This is a telescoping sum,so $I = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

Let $n$ be a positive integer. For a real number $x$,let $[x]$ denote the largest integer not exceeding $x$ and $\{x\}=x-[x]$. Then,$\int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$ is equal to

Similar Questions

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

$\int_{0}^{1} \tan^{-1}\left(\frac{2x}{1-x^2}\right) dx =$

If $\int_0^{k} \frac{d x}{2+8 x^2}=\frac{\pi}{16}$,then the value of $k$ is

If $I$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$, then

$\int_0^1 \sqrt{\frac{1-x}{1+x}} \,dx$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module