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DOUBLE INTEGRATE: Ouside Integal (v=1, v=4) Inside Integral (u=0, u=1) [ ve^(uv)dudv ] ... i dont kno what to do with du??
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\[\int_1^4\int_0^1ve^{uv}dudv\]?
yes
wait there is also (1/2) in the beginning
do this first, treat \(v\) as a constant \[\int_0^1ve^{uv}du\]
so e^uv , where e^v - 1 will be left
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yes that is what i get
then \[\int_1^4(e^v-1)dv=\int_1^4e^vdv-3\] etc
ok got it it was easy
yeah, just hope it was right!
(1/2)(e^4 - e - 3) how? that is the answer
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that is what i get, assuming you have a \(\frac{1}{2}\) outside
ok
anti derivative of \(e^v\) is \(e^v\) plug in 4 get \(e^4\) plug in 1 get \(e\) subtract get \(e^4-e\)
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