If n is a positive integer and [x] is the gre | vedclass

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

Question

(C) The expression ${x - [x]}$ represents the fractional part of $x$,denoted as $\{x\}$.Since the fractional part function $f(x) = \{x\} = x - [x]$ is

Vedclass · Accepted Answer

$n/2$

Vedclass · Answer

(C) The expression ${x - [x]}$ represents the fractional part of $x$,denoted as $\{x\}$.Since the fractional part function $f(x) = \{x\} = x - [x]$ is a periodic function with period $T = 1$,we can use the property $\int_0^{nT} f(x) \,dx = n \int_0^T f(x) \,dx$.Therefore,$\int_0^n \{x\} \,dx = n \int_0^1 \{x\} \,dx$.In the interval $[0, 1)$,$[x] = 0$,so $\{x\} = x - 0 = x$.Thus,$\int_0^n \{x\} \,dx = n \int_0^1 x \,dx$.Evaluating the integral: $n \left[ \frac{x^2}{2} \right]_0^1 = n \left( \frac{1}{2} - 0 \right) = \frac{n}{2}$.

If $n$ is a positive integer and $[x]$ is the greatest integer not exceeding $x$,then $\int_0^n {\{x - [x]\} \,dx}$ equals

Similar Questions

$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx = \dots$

The option$(s)$ with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4 \pi} e^t(\sin^6 at + \cos^4 at) dt}{\int_0^{\pi} e^t(\sin^6 at + \cos^4 at) dt} = L$.

$\int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$ is equal to:

Let $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$ for all $x \in [0, 1]$,$p(0) = 1$,and $p(1) = 41$. Then $\int_{0}^{1} p(x) dx = $

If $f(t) = \int_0^t \tan^{(2n-1)} x \, dx$,$n \in N$,then $f(t+\pi) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module