# Improper integral of integral test

1. Sep 20, 2010

### ProPatto16

1. The problem statement, all variables and given/known data

determine the value of the improper integral when using the integral test to show that $$\sum$$ k / e^k/5 is convergent.

a) 50/e
b) -1 / 5e^1/5
c) 5
d) 5e
e)1/50e

3. The attempt at a solution

f(x) = xe^-x/5 is continuos and positive for all values x>0
to see where its decreasing, take derivitive and see where f'(x) < 0

f'(x) = -x/5*e^-x/5 + e^-1/5
this function is < 0 only when x>5... so f(x) is continuous, positive and decreasing when x>5... therefore lower limit is 5.

$$\int$$f(x).dx = $$\int$$ xe^-x/5 .dx

= [-x/5*e^-x/5 + e^-1/5] between b and 5 where b replaces infinity.

* Note here that f'(x) = $$\int$$ f(x)... im not sure why but thats what i got.

if i sub in the lower bound 5 i get -e^-1-e^-1

which doesnt match anything :S

stuckkkkkk.

2. Sep 20, 2010

### Staff: Mentor

This is ambiguous. Is the function the first or the second?
$$\frac{k}{\frac{e^k}{5}}$$

or
$$\frac{k}{e^{k/5}}$$

3. Sep 20, 2010

### snipez90

OK fine, 5 is the lower limit of integration. It turns out that it doesn't really matter in this case, so a problem like this is really somewhat against the spirit of analysis.

Again, there are problems with your integration that go beyond terminology. First of all, assuming the first term is your 'uv term', it's wrong because when you integrate e^(-x/5), you shouldn't get the -1/5 out front. Note that when you differentiate e^(-x/5), you do in fact get (-1/5)e^(-x/5), so do you see now how to integrate e^(-x/5)?

Secondly, please show the integration of v*du because I don't think what you have there is correct.

4. Sep 20, 2010

### ProPatto16

the function, mark,is the second one you have there. e^(k/5).. ill get back to this one soon, im still fiddling with the other one you guys are helpin me with. thanksss

5. Sep 21, 2010

### ProPatto16

hey guys, i re calculated the integral using product rule and substitution.... i got a differnet answer, ill show you..

product rule: (uv)'=uv'+vu'
f(x)=xe^(-x/5)
let u=x therefore u'=1
let v=e^(-x/5) and use subsititution to find v'

v=e^(-x/5), let u=(-x/5) therefore v=e^u so v'=e^u=e^(-x/5)

then $$\int$$ f(x).dx= xe^(-x/5) + e^(-x/5)

im sure the maths is right there im just not sure about that + sign and whether it should be a -... should my substitution be u=x/5 then have e^-u?? so then derivitive is -e^(-x/5)

after that im not sure how to incorporate the bounds when the ubber bound is b (substituted for infinity) and the lower bound is 5.

6. Sep 21, 2010

### ProPatto16

no wait, thats wrong.

v'=-5e^(-x/5).... i forgot to do anything with my substitution of u

so $$\sum$$ f(x).dx = -5xe^(-x/5) + e^(-x/5)

im sure thats right. and dont worry about the u part in the previous post... it works out the same with or without the negative.

Last edited: Sep 21, 2010
7. Sep 21, 2010

### ProPatto16

Thats all wrong.... i derived -.- so dumb

using by parts..

$$\int$$ f(x)g'(x) .dx = f(x)g(x) - $$\int$$ g(x)f'(x)

let f(x) = x, so f'(x) = 1
and g'(x)= e(-x/5) so g(x)=-5e(-x/5)

so $$\int$$ xe(-x/5) .dx = x*-5e(-x/5) - $$\int$$ -5e(-x/5)*1

= -5xe(-x/5) - e(-x/5)

there.

Last edited: Sep 21, 2010
8. Sep 22, 2010

### ProPatto16

so thats the integral... but i dont know how to get from there to "the value of the improper integral", which is what i need for the answer.

9. Sep 22, 2010

### snipez90

Good, I think that looks right. Now for the first term, call it h(x) = (-5)xe^(-x/5), you need to take the difference h(b) - h(5) and take the limit of this as b goes to infinity. Here, recall that the exponential function dominates polynomials.

For the second term (the integral) do something similar. Evaluate the integral with lower limit of integration 5 and upper limit b, and let b go to infinity.

10. Sep 22, 2010

### ProPatto16

As b -> infinity, e^(-b/5) is 0. so all b terms become zero... so:

lim [-5xe^(-x/5) - e^(-x/5)] with upper bound b and lower bound 5 becomes

lim [(0 - 0) - (-25e^(-1) - e^(-1)] which becomes

lim [0 - 0 + 25e^(-1) + e^(-1)]

.....

11. Sep 22, 2010

### ProPatto16

and if i do it seperately....

take first term, -5xe^(-x/5) :
= (-5b^(-b/5) - -5(5)^e(-5/5))
= (0 + 25e^(-1))
= 25e^(-1)

second term, -e^(-x/5)
= -e^(-b/5) - -e^(-5/5)
= 0 + e^(-1)
= e^(-1)

....

12. Sep 22, 2010

### snipez90

Almost, you have a 5 in front of the integral to start with. What is the integral of e^(-x/5) again?

13. Sep 22, 2010

### ProPatto16

integral of e^(-x/5) is

-5e^(-x/5)

but why am i integrating again? i can see there might be a little confusion where i did my integral... i skipped a step near the end... but the complete integral of the given function is

-5xe(-x/5) - e(-x/5)

so howcome i need to integrate the second term?

14. Sep 22, 2010

### ProPatto16

ohhh the by parts integration is wrong towards the end.

so $$\int$$ xe(-x/5) .dx = x*-5e(-x/5) - $$\int$$ -5e(-x/5)*1

and that second integral there is $$\int$$ -5e(-x/5)*1

= -5 $$\int$$ e^(-x/5)
= -5 * -5e(-x/5)
= 25e^(-x/5)

so the complete integral should be:

= -5xe(-x/5) - 25e(-x/5)

and then somewhere in the maths ive gone wrong with the signs... cause then if they're both +ve instead of -ve... the answer is 50e^(-1)

....

15. Sep 23, 2010

### Staff: Mentor

Nowhere in this thread do I see a definite, improper integral. To determine the convergence of the given series, evaluate this integral.
$$\int_1^{\infty} xe^{-x/5}dx$$

Since this is an improper integral, you need to evaluate this limit to determine the value of the integral.
$$\lim_{b \to \infty}\int_1^b xe^{-x/5}dx$$

Click either of the expressions above to see my LaTeX code.