Posted by Katie on Sunday, December 30, 2012 at 5:23pm.
Integrate using integration by parts
(integral) (5-x) e^3x
u = 5-x
du = -dx
dv = e^3x
v = 3e^3x
I wonder if this is right so far.
= uv - (integral) v du
= (5-x)(3e^3x) - (integral) (-3e^3x)
=(5-x)(3e^3x) + (integral) (3e^3x)
= (5-x)(3e^3x) + 9e^3x + C
Calculus II - Damon, Sunday, December 30, 2012 at 5:36pm
dv = e^3x
v = (1/3)e^3x !!!!!!
Calculus II - Katie, Sunday, December 30, 2012 at 6:34pm
Okay then, is this the final answer
= (5-x)(3e^3x) + (integral) -- (1/3)e^3x
= (5-x)(3e^3x) + (1/9)e^3x + C
Calculus II - drwls, Sunday, December 30, 2012 at 6:47pm
No. You are using the wrong v function in the u*v term. See Damon's answer.
The (Integral)-v*du term is:
So your answer is ok for that term
Calculus II - Katie, Sunday, December 30, 2012 at 7:02pm
ok, thank you very much. Happy New Year
Calculus II - jaime, Sunday, September 13, 2015 at 11:55pm
Calculus II - jaime, Sunday, September 13, 2015 at 11:57pm
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