The value of \(\int_{0}^{1} { e }^{ -2x } dx\)1\[\frac{{ e }^{ -2} -1}{-2}\]2\[\frac{{ e }^{- 2 } -1}{2}\]3\[\frac{-{ e }^{ -2 } +1}{-2}\]4\[\frac{-{ e }^{ 2 } +1}{2}\]Answer & explanation+Correct answer: option 1the integral will be \[\frac{{ e }^{ -2x } }{-2}\]
