Integral Calculus

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Question : 62 of 73

Marks: +1, -0

∫ e^{-3 x} d x=

${1}/{3} e^{-3 x}+C$
$\;{1}/{3} e^{-3 x}+c$
${1}/{3} e^{3 x}+c$
$-3 e^{-3 x}+c$

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