Given$$f(x)=e ^{-x}(x-1)$$.
(a) find f'(x).
(b) use your result from (a) to find $$\int\limits xe^{-x}$$

Help please. I have done part (a) but don't know what to do in part (b). Can anyone help me please.

Thanks.

Question

Given$$f(x)=e ^{-x}(x-1)$$. 
(a) find f'(x).
(b) use your result from (a) to find $$\int\limits xe^{-x}$$

Help please. I have done part (a) but don't know what to do in part (b). Can anyone help me please. 

Thanks.

rahulmr · Answer

this is what i got for part (a) $$2e^{-x}-e^{-x}x$$

zepdrix · Answer

Your derivative looks good.
And of course we could solve this integral using Integration-by-parts.
But that's not what they want here...

I'm trying to see how they're connecting these two things together...
If you factor your expression, it might become a little more clear,
$\large\rm -e^{-x}(x-2)$

rahulmr · Answer

hmm.....what do you think i should i do?

zepdrix · Answer

Well notice what happened,
you lost another 1 from your x,
and also the sign changed.

Interesting to note,
if you took another derivative, you would get,$$\large\rm e^{-x}(x-3)$$Losing another 1 from the x, and the sign changes again, ya?

legomyego180 · Answer

what a weird question

zepdrix · Answer

Note: Derivative is causing us to `lose 1` each time.
So integration will cause us to `gain 1`, ya?

So if we start with this,$$\large\rm e^{-x}(x-0)$$(which is what our expression is)
do you see how we can apply this pattern?

rahulmr · Answer

yes.

rahulmr · Answer

but i'm not sure how this will relate to the (b)

zepdrix · Answer

Differentation causes us to `lose 1` and `sign change`.
So Integration should cause us to `gain 1` and `sign change`.

zepdrix · Answer

Example, if we started with $\large\rm -e^{-x}(x-7)$
Integrating should give us $\large\rm e^{-x}(x-6)$

rahulmr · Answer

Oh, Now i get it.

zepdrix · Answer

So what do you end up with after integrating? :)

rahulmr · Answer

$$(-x-1)e^{-x}$$

rahulmr · Answer

is it correct ?

zepdrix · Answer

I'm not sure why you wrote it like that... that's kinda confusing given the pattern we were trying to follow. hmm :p

zepdrix · Answer

In part b, they gave us this,$$\large\rm \int\limits e^{-x}(x-0)dx$$So we'll change the sign, and add 1 to our x,$$\large\rm -e^{-x}(x+1)$$Which yes, is equivalent to your expression.

I just don't understand why yours looks like that.
Is it from an answer key or something? +_+

rahulmr · Answer

sorry. did you meant like this $$-e^{-x}(x+1)$$

rahulmr · Answer

Thanks a lot for the help @zepdrix

zepdrix · Answer

yay team \c:/