For each real number x,let [x] denote the gre | vedclass

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

Question

(C) We are given the integral $I = \int_1^M \{x\}^{[x]} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can write the integral as a sum of int

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) We are given the integral $I = \int_1^M \{x\}^{[x]} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can write the integral as a sum of integrals over unit intervals: $I = \sum_{n=1}^{M-1} \int_n^{n+1} (x-n)^n dx$. Let $u = x-n$,then $du = dx$. The integral becomes $\int_0^1 u^n du = \left[ \frac{u^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}$. Thus,$I = \sum_{n=1}^{M-1} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{M}$. For $M=2$,$I = \frac{1}{2} = 0.5 For $M=3$,$I = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \approx 0.833 For $M=4$,$I = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6+4+3}{12} = \frac{13}{12} > 1$. Therefore,the smallest positive integer $M$ is $4$.

For each real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$,and let $\{x\} = x - [x]$. Then the smallest positive integer $M$ for which $\int_1^M \{x\}^{[x]} dx > 1$ is

Similar Questions

The greatest integer less than or equal to $\int_1^2 \log _2(x^3+1) dx + \int_1^{\log_2 9} (2^x-1)^{1/3} dx$ is . . . . .

If $f(x)$ satisfies the relation $f(x) = e^{x} + \int_{0}^{1} (y + xe^{x}) f(y) dy$, then $e + f(0)$ is equal to . . . . . . .

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

If $f(x) = A \sin \left( \frac{\pi x}{2} \right) + B$,$f'(1/2) = \sqrt{2}$ and $\int_{0}^{1} f(x) dx = \frac{2A}{\pi}$,then the constants $A$ and $B$ are respectively.

Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module