1. ## integration of parts

Hi,

Can someone please tell me if this is correct? I'm trying to solve this integral with the integration of parts method.

The integral xe^(-x)dx

udv=uv-the integral vdu

u=x
dv=e^-xdx

du=1dx (derivative of u)
v=-e^-x (antiderivative of du)

xe^-x-the integral -e^-x=
e^-x+e^-x=
e^-x(x+1)+c

Thank you very much

2. Originally Posted by chocolatelover
Hi,

Can someone please tell me if this is correct? I'm trying to solve this integral with the integration of parts method.

The integral xe^(-x)dx

udv=uv-the integral vdu

u=x
dv=e^-xdx

du=1dx (derivative of u)
v=-e^-x (antiderivative of du)

xe^-x-the integral -e^-x=
e^-x+e^-x=
e^-x(x+1)+c

Thank you very much
You need to sort out how you are expressing things in ASCII.

You should have:

int xe^(-x)dx = (-x e^{-x}) + int e^{-x} dx = -x e^{-x} - e^{-x} + C

which you can now simplify, and check by differentiating

RonL

3. Hello, chocolatelover!

You dropped a minus-sign . . .

Is this correct?

Integrate: . $\int xe^{\text{-}x}dx$

By parts: . $\int u\,dv\;=\;uv - \int v\,du$

. $\begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{\text{-}x}dx \\

du & = & dx & \quad & v & = & \text{-}e^{\text{-}x}\end{array}$

Then we have: . $-xe^{\text{-}x} + \int e^{\text{-}x}dx \;=\;-xe^{\text{-}x} - e^{\text{-}x} + C \;=\;-e^{\text{-}x}(x +1) + C$

4. Hi,

I looked at it again. What would the antiderivative of -e^-x be? Is that just "-e^-x"?

Thank you

5. Originally Posted by chocolatelover
Hi,

I looked at it again. What would the antiderivative of -e^-x be? Is that just "-e^-x"?

Thank you
Technically it's $\frac{-e^{-x}}{-1} = e^{-x}$. So not quite.

In general $\frac{d}{dx}e^{kx} = ke^x$. So $\int e^{kx}~dx = \frac{e^{kx}}{k} + C$.

-Dan