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

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(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 $(G.I.F.)$.

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