int_{-10}^{10} frac{3^x}{3^{[x]}} , dx is equ | vedclass

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

Question

(B) We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function.Thus,$\frac{3^x}{3^{[x]}} = 3^{x-[x]} = 3^{\{x\}}$.The integral becom

Vedclass · Accepted Answer

$\frac{40}{\ln 3}$

Vedclass · Answer

(B) We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part function.Thus,$\frac{3^x}{3^{[x]}} = 3^{x-[x]} = 3^{\{x\}}$.The integral becomes $I = \int_{-10}^{10} 3^{\{x\}} \, dx$.Since $\{x\}$ is a periodic function with period $1$,and the interval length is $20$,we have:$I = 20 \int_{0}^{1} 3^x \, dx$.Evaluating the integral:$I = 20 \left[ \frac{3^x}{\ln 3} \right]_{0}^{1} = \frac{20}{\ln 3} (3^1 - 3^0) = \frac{20}{\ln 3} (3 - 1) = \frac{40}{\ln 3}$.

$\int_{-10}^{10} \frac{3^x}{3^{[x]}} \, dx$ is equal to,where $[ \cdot ]$ denotes the Greatest Integer Function $(G.I.F.)$.

Similar Questions

Let the function $F$ be defined as $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt, x > 0$. Then the value of the integral $\int_{1}^{x} \frac{e^{t}}{t+a} dt$,where $a > 0$,is

$\int_{0}^{1} x(1 - x)^{5} dx = . . . . . .$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x=$

If $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \\ 2, & \text{otherwise} \end{cases}$,then $\int_{-2}^{3} f(x) dx$ is equal to

$\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module