## find the antiderivative of x^2e^(3x)

Limits, differentiation, related rates, integration, trig integrals, etc.
Martingale
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### Re: find the antiderivative of x^2e^(3x)

for the antiderivative of the second part then I would get

(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)

$=\frac{2x}{3}\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}\frac{2}{3}dx=\frac{2xe^{3x}}{9}-\int\frac{2e^{3x}}{9}dx$

yabo2k
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### Re: find the antiderivative of x^2e^(3x)

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

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### Re: find the antiderivative of x^2e^(3x)

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
for the antiderivative of the second part then I would get

(2x/3)(e^3x/3)-(e^3x/3)((2/3)dx)

$=\frac{2x}{3}\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}\frac{2}{3}dx=\frac{2xe^{3x}}{9}-\int\frac{2e^{3x}}{9}dx$
$=\frac{2x}{3}\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}\frac{2}{3}dx=\frac{2xe^{3x}}{9}-\int\frac{2e^{3x}}{9}dx=\frac{2xe^{3x}}{9}-\frac{2e^{3x}}{27}+c$

yabo2k
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### Re: find the antiderivative of x^2e^(3x)

so did you find u, du, dv, and v again to get (2e^3x)/27

Martingale
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### Re: find the antiderivative of x^2e^(3x)

so did you find u, du, dv, and v again to get (2e^3x)/27
no...not for the last part. just used the fact that $\int e^{3x}dx=e^{3x}/3+c$ one last time

yabo2k
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### Re: find the antiderivative of x^2e^(3x)

so my final answer would be

((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C

Martingale
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### Re: find the antiderivative of x^2e^(3x)

so my final answer would be

((x^2e^3x)/3)-((2xe^3x)/9)-((2e^3x)/27)+C
almost..check the sign on the (2e^3x)/27 term

yabo2k
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### Re: find the antiderivative of x^2e^(3x)

$=\frac{2x}{3}\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}\frac{2}{3}dx=\frac{2xe^{3x}}{9}-\int\frac{2e^{3x}}{9}dx=\frac{2xe^{3x}}{9}-\frac{2e^{3x}}{27}+c$[/quote]

You have it as a subtraction. So where did i go wrong?

Martingale
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### Re: find the antiderivative of x^2e^(3x)

$=\frac{2x}{3}\frac{e^{3x}}{3}-\int\frac{e^{3x}}{3}\frac{2}{3}dx=\frac{2xe^{3x}}{9}-\int\frac{2e^{3x}}{9}dx=\frac{2xe^{3x}}{9}-\frac{2e^{3x}}{27}+c$

You have it as a subtraction. So where did i go wrong?[/quote]

I have a subtraction for this small part of the problem. Combine this with the full problem.

Martingale
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### Re: find the antiderivative of x^2e^(3x)

did you see this post of mine?