Let f(x) = x - [x], for every real number x, | vedclass

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

Question

(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.For the interval $[-1, 1]$,we split the integral at $x = 0$:$\int

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.For the interval $[-1, 1]$,we split the integral at $x = 0$:$\int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx$.For $-1 \le x For $0 \le x Now,calculate the integrals:$\int_{-1}^{0} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = (0 + 0) - \left( \frac{1}{2} - 1 \right) = -(-1/2) = 1/2$.$\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 1/2$.Adding these results: $1/2 + 1/2 = 1$.

Let $f(x) = x - [x],$ for every real number $x,$ where $[x]$ is the integral part of $x.$ Then $\int_{-1}^{1} f(x) \, dx =$

Similar Questions

$\int_{-2}^2 |[x]| \, dx$ is equal to

If $\int_0^1 {x \log \left( {1 + \frac{x}{2}} \right)} \,dx = a + b \log \frac{2}{3},$ then

If $[\bullet]$ denotes the greatest integer function, then evaluate $\int_0^{2 \pi} [|\sin x| + |\cos x|] \, dx$.

$\int_0^{\pi / 4} \tan ^2(x) \, dx =$

Let $f(x) = \begin{cases} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{cases}$. Consider the integral $I_n = \sqrt{n} \int_0^{1/n} f(x) e^{-nx} dx$. Then,$\lim_{n \to \infty} I_n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module