Let I(R) = int_0^R e^{-R sin x} dx, where R | vedclass

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(D) Consider the function $f(x) = \sin x$. On the interval $[0, \pi/2]$, we know that $\sin x \geq \frac{2x}{\pi}$ by Jordan's inequality.Since $R > 0

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$I(R)$ and $\frac{\pi}{2R}(1 - e^{-R})$ are not comparable

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(D) Consider the function $f(x) = \sin x$. On the interval $[0, \pi/2]$, we know that $\sin x \geq \frac{2x}{\pi}$ by Jordan's inequality.Since $R > 0$, we have $-R \sin x \leq -R \frac{2x}{\pi}$.Thus, $e^{-R \sin x} \leq e^{-2Rx/\pi}$.Integrating both sides from $0$ to $\pi/2$ gives $\int_0^{\pi/2} e^{-R \sin x} dx \leq \int_0^{\pi/2} e^{-2Rx/\pi} dx = \frac{\pi}{2R}(1 - e^{-R})$.However, the integral $I(R)$ is defined from $0$ to $R$. For large $R$, the behavior of the integral is dominated by the region near $x=0$ where $\sin x \approx x$.By comparing the growth rates and the bounds of the integral, it is observed that for general $R > 0$, the values of $I(R)$ and the expression $\frac{\pi}{2R}(1 - e^{-R})$ do not maintain a consistent inequality for all $R$, making them not directly comparable in a simple global sense.

Let $I(R) = \int_0^R e^{-R \sin x} dx$, where $R > 0$. Then,

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