I know how to solve but the definite integral to 1 to infinite has thrown me off. Thank you for all the help.
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int (x * e^(-3x)) dx
= (-1/9)(3x + 1) e^(-3x) + C
sub in limits
= (-1/9)(3∞ + 1) e^(-3∞) - (-1/9)(3(1) + 1) e^(-3(1))
with the first group
the limit as x -> ∞ of e^(-3x) is 0
therefore,
= 0 + (1/9)(4) e^-3
= (4/9) e^-3
= (-1/9)(3x + 1) e^(-3x) + C
sub in limits
= (-1/9)(3∞ + 1) e^(-3∞) - (-1/9)(3(1) + 1) e^(-3(1))
with the first group
the limit as x -> ∞ of e^(-3x) is 0
therefore,
= 0 + (1/9)(4) e^-3
= (4/9) e^-3
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That problem just looks like random symbols, Idk how the heck to figure that out