The value of \(\int { e }^{ -8x } dx\)

\[\frac{{ e }^{ -8x }}{8}\] \[\frac{{ e }^{ -8x }}{-8}\] \[\frac{{ e }^{ 8x }}{8}\] \[\frac{{ e }^{ 8x }}{-8}\]

\[\frac{{ e }^{ -8x }}{-8}\]

the integral will be \[\frac{{ e }^{-8x} }{-8}\]