∫ x * e^(-x) dx = ?

Question

I'm finding myself stumped by this one, as I keep getting the wrong answer....how do you go about solving:

∫ x * e^(-x) dx = ?

Many thanks!

Ty guys for the help so far....but I'm still not getting the first step you guys are showing me...how can you justify splitting ∫ x * e^(-x) dx into two terms at the start to get - x e^(-x) + ∫ e^(-x) d x ??

In other words, what rule is there to let you do that? (I think I need the first step kind of spelled out for me here :( )

Answer 1

∫xe^(-x) dx

We proceed with integration by parts: let u=x, dv=e^(-x) dx, du=dx, v=-e^(-x). Then we have:

-xe^(-x) - ∫-e^(-x) dx
-xe^(-x) + ∫e^(-x) dx
-xe^(-x) - e^(-x) + C

And we are done.

Edit: the rule is called integration by parts. The rule is this: Suppose u and v are differentiable functions of x, then:

∫u dv = uv - ∫v du

This is derived from the product rule for differentiation as follows:

(uv)' = uv' + u'v
uv' = (uv)' - u'v
Integrate both sides:
∫uv' = uv - ∫u'v

Which (with appropriate change of notation) is exactly what I wrote above.

Integration by parts is useful whenever you are asked to integrate a product of two functions. I indicated explicitly which function was interpreted as u and which function was interpreted as dv for the purpose of invoking this rule. Note that the choice of functions here is completely arbitrary - I could have chosen, say, u=e^(-x) and dv=x, which would yield:

∫xe^(-x) dx = x²e^(-x)/2 - ∫-x²e^(-x)/2 dx

Which is completely correct. However I did not do this, for the simple reason that it obviously does not help us solve the problem. Learning when to use integration by parts, and how, is somewhat more art than science, but luckily you'll get plenty of practice in your calculus class doing precisely that (cue moaning).

Answer 2

∫ x * e^(-x) dx is found by integration by parts;

∫ x e^(-x) dx = - ∫ x d {e^(-x) }
= - x e^(-x) + ∫ e^(-x) d x
= - x e^(-x) - e^(-x) + C
= - e^(-x) ( x + 1) + C