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

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$, then $b$ is equal to

If $u(n) = \int_0^{\frac{\pi}{2}} (1 + \sin t)^n \sin 2t \, dt$,where $n \in N$,then $u(4) = $

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$

If $I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$,then $I_{n-1} + I_{n+1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module