Let f(x) = max {x+|x|, x-[x]}, where [x] deno | vedclass

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

Question

(C) Given, $f(x) = \max \{x+|x|, x-[x]\}$.For $x \geq 0$, $x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$, $f(x) = 2

Vedclass · Accepted Answer

$21/2$

Vedclass · Answer

(C) Given, $f(x) = \max \{x+|x|, x-[x]\}$.For $x \geq 0$, $x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$, $f(x) = 2x$.For $x Thus, $\int_{-3}^{3} f(x) dx = \int_{-3}^{0} (x-[x]) dx + \int_{0}^{3} 2x dx$.Since $x-[x]$ is periodic with period $1$, $\int_{-3}^{0} (x-[x]) dx = 3 \int_{0}^{1} x dx = 3 \left[ \frac{x^2}{2} \right]_0^1 = \frac{3}{2}$.And $\int_{0}^{3} 2x dx = [x^2]_0^3 = 9$.Therefore, $\int_{-3}^{3} f(x) dx = \frac{3}{2} + 9 = \frac{21}{2}$.

Let $f(x) = \max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) dx$ is

Similar Questions

$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $

Evaluate the definite integral $\int_{0}^{9} [\sqrt{x} + 2] \, dx$,where $[\cdot]$ denotes the Greatest Integer Function $(G.I.F.)$.

$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{1 + {x^2}}}} \right)} \,dx$ is equal to

$\int_5^{10} \frac{d x}{(x-1)(x-2)} = $

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$, then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module