Let {\I}( {\R})=\int_{0}^{ {\R}} {\e}^{- {\R} \sin x} {\d} x, {\R}>0. then,

Correct Answer is : I(R) and π2R(1−e−R) are not comparable

\I(\R)>\frac{\pi}{2 R} (1-e^{-R})

\I(\R)< \frac{\pi}{2 R} (1-e^{-R})

\I(\R)=\frac{\pi}{2 R} (1-e^{-R})

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

WBJEE 2024 Math Paper

Question : 51 of 75

Marks: +1, -0

Let I(R)=

∫

e−Rsin‌xdx,R>0. then,

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