Integration by Parts: Solve \int (xe^-^x)dx

[suspenc3]

hi..im new to this topic..can someone check to see if this is right?

$$\int (xe^-^x)dx = \int udV = uV - \int Vdu$$
$$=x(-e^-^x)- \int -e^-^x$$
$$=-xe^-^x-e^-^x+C$$

thanks

 

[arildno]

Why not check it yourself?
Differentiate your answer and see if you get back your integrand!

 

[suspenc3]

wow..why didnt i think of that haha

THanks

 

[arildno]

Spoiler alert!




Yeah, it's right

 

[suspenc3]

Ok, I have another one I am trying to figure out...but i keep getting 0!

$$\int_{0}^{\pi} tsin3t dt = udV \right]_{0}^{\pi} - \int_{0}^{\pi}vdu$$

work...

$$=t(sin3t) \\right]_{0}^{\pi} - \int_{0}^{\pi}vdu$$

$$=\pi (sin3\pi) - \int_{0}^{\pi} \frac{-1}{3} cos3t(1)dt$$

$$=0 - \frac{1}{sin3t} \int_{0}^{\pi}z dz$$..
let z = cos3t
and then use substitution to get $$\frac{dz}{sin3t} = -3dt$$

$$0-\frac{1}{sin3t} \frac{z^2}{2}$$

$$=\frac{-(cos3t)^2}{2sin3t}$$
this must be wrong..i solved it out a bit more..but keep getting zero, where did i go wrong?

 

[arildno]

Well, what you have written is just nonsense.
As a novice, it is safer for you to do these problems like this:
$$u(t)=t\to\frac{du}{dt}=1,\frac{dv}{dt}=\sin(3t)\to{v}(t)=-\frac{1}{3}\cos(3t)$$

Use these relations in the integration by parts formula.

 

[suspenc3]

hrmm..i already knew these relations..i still don't see what i did wrong..i followed it just how they do it in the book, my LaTeX skills arent very good, is it just hard to follow?

 

[arildno]

well, you didn't compute the uv-term correctly.

 

[suspenc3]

o wait..it is :$$uv - \int_{0}^{\pi}vdu$$
i have $$udv - \int_{0}^{\pi}vdu$$ in my work..my bad

 

[suspenc3]

giving me $$\frac{\pi}{3}$$ which i think is right..thanks for the help

 

[HallsofIvy]

I would strongly recommend that you write out exactly what "u", "v", "du", and "dv" are!