Let {x} denote the fractional part of a real | vedclass

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

Question

(B) The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$.For the interval $[0, 2)$, we have:$\{x\} = x$ for $0 \leq x <

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$.For the interval $[0, 2)$, we have:$\{x\} = x$ for $0 \leq x $\{x\} = x - 1$ for $1 \leq x Therefore, the integral can be split as:$\int_0^2 \{x\} dx = \int_0^1 x dx + \int_1^2 (x - 1) dx$Evaluating the first part: $\int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$Evaluating the second part: $\int_1^2 (x - 1) dx$. Let $u = x - 1$, then $du = dx$. When $x=1, u=0$; when $x=2, u=1$.$\int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$Adding both parts: $\frac{1}{2} + \frac{1}{2} = 1$.

Let $\{x\}$ denote the fractional part of a real number $x$. Then $\int_0^2 \{x\} dx = $

Similar Questions

$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $

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

$\int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx = } $

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

$\int_0^1 {(1 - x)^9} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module