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Equation posted inside. calculus integral
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give me a min to get the equation in
\[\int\limits 6re^r/3 dr\]
i can post what i have so far but I will need a min for that too. Jordan are you here?
By parts, regardless of whether you meant \[\int6re^{r/3}\,dr\quad\text{or}\quad\int6r\frac{e^r}{3}\,dr\]
i meant the first one you wrote
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u= 6r dv= e^r/3 du= 6dr w= r/3 dr= 3dw v= 3e^ r/3
6r (3e^r/3) - integral 3 e^r/3 (6dr) = 18 re^r/3 - 18 integral e^r/5 dr
this is where i get stuck
\[18 r e^\frac{ r }{ 3 } - 18 \int\limits e^\frac{ r }{ 5 } dr\]
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