The value of int_{-7}^{7} frac{5^x}{5^{[x]}} | vedclass

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Question

(C) Let $I = \int_{-7}^{7} \frac{5^x}{5^{[x]}} dx$. Since $x - [x] = \{x\}$ (fractional part of $x$),the integral becomes $I = \int_{-7}^{7} 5^{\{x\}}

Vedclass · Accepted Answer

$\frac{56}{\ln 5}$

Vedclass · Answer

(C) Let $I = \int_{-7}^{7} \frac{5^x}{5^{[x]}} dx$. Since $x - [x] = \{x\}$ (fractional part of $x$),the integral becomes $I = \int_{-7}^{7} 5^{\{x\}} dx$. Because the function $f(x) = 5^{\{x\}}$ is periodic with period $T = 1$,we can write: $I = 14 \int_{0}^{1} 5^x dx$. Evaluating the integral: $I = 14 \left[ \frac{5^x}{\ln 5} ight]_{0}^{1}$. $I = \frac{14}{\ln 5} (5^1 - 5^0) = \frac{14}{\ln 5} (5 - 1)$. $I = \frac{14 	imes 4}{\ln 5} = \frac{56}{\ln 5}$.

The value of $\int_{-7}^{7} \frac{5^x}{5^{[x]}} dx$ is equal to (where $[.]$ denotes the greatest integer function).

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