Let [x] and {x} be the integer part and fract | vedclass

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

Question

(B) Let $I = \int_0^5 [x]\{x\} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can split the integral: $I = \sum_{n=0}^{4} \int_n^{n+1} [x]\{x

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $I = \int_0^5 [x]\{x\} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can split the integral: $I = \sum_{n=0}^{4} \int_n^{n+1} [x]\{x\} dx$. For $x \in [n, n+1)$,$[x] = n$ and $\{x\} = x - n$. Thus,$I = \sum_{n=0}^{4} \int_n^{n+1} n(x-n) dx$. Let $t = x-n$,then $dt = dx$. When $x=n, t=0$ and when $x=n+1, t=1$. $I = \sum_{n=0}^{4} n \int_0^1 t dt = \sum_{n=0}^{4} n \left[ \frac{t^2}{2} \right]_0^1 = \sum_{n=0}^{4} n \left( \frac{1}{2} \right) = \frac{1}{2} (0+1+2+3+4) = \frac{10}{2} = 5$.

Let $[x]$ and $\{x\}$ be the integer part and fractional part of a real number $x$ respectively. The value of the integral $\int_0^5 [x]\{x\} dx$ is

Similar Questions

$\int_{0}^{\pi} \sin x \, dx = $ . . . . . . .

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

$\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x} = $

$\int_0^{1.5} {[x^2] \, dx}$,where $[.]$ denotes the greatest integer function,equals

If $x({x^4} + 1)\phi (x) = 1,$ then $\int_1^2 {\phi (x)\,dx = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module