yabo2k wrote:for the antiderivative of the second part then I would get
(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)
you really need to put an integral sign in there
yabo2k wrote:for the antiderivative of the second part then I would get
(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)
yabo2k wrote:I have them on my paper but I don't know how to put it into the computer. But I guess this is where I get stuck. I don't see how it can go any further than my first answer
Martingale wrote:yabo2k wrote:for the antiderivative of the second part then I would get
(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)
you really need to put an integral sign in there
yabo2k wrote:so did you find u, du, dv, and v again to get (2e^3x)/27
yabo2k wrote:so my final answer would be
((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C
yabo2k wrote: