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

The value$(s)$ of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

The value of $\int_{0}^{\pi /2} (\sqrt{\sin \theta} \cos \theta)^3 d\theta$ is

Evaluate the definite integral $\int_{0}^{1} \frac{d x}{1+x^{2}}$.

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

$\int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module