Question

"Calculate the following improper integrals. If the integral diverges, show why".. I'm struggling to understand this last part of the question. What does one find if an integral does diverge and how does one go about showing it if so?

Answer 1

what integral?

Answer 2

If the limit does not exist or it is infinite, then we say that the improper integral is divergent.

Answer 3

There's three that are given:
\[\int\limits_{1}^{\infty}g^{-4}dg\]
\[\int\limits_{1}^{2}\frac{ 1 }{ (2-1)^{2} }ds\]
\[\int\limits_{0}^{\infty}\frac{ -2t }{e ^{2t} }dt\]

Answer 4

take the anti derivative, then compute the limit

Answer 5

Ah, so the second integral, which I found to equal -1/3 + 1/3(infinity) or just infinity, is divergent?

Answer 6

the second integral has a typo in it

Answer 7

apologies, the 2 should be a s, yes

Answer 8

In the denominator

Answer 9

did you find the anti - derivative?

Answer 10

i get
\[\frac{1}{1-x}\] at \(2\) you have no problem but at \(1\) you are out of luck

Answer 11

if you want to be more formal
\[\lim_{l\to 1}\frac{1}{1-l}\] does not exist

Answer 12

Perfect - thank you for your help with that!

Answer 13

you good with the others?

Answer 14

\[\int x^{-4}dx=-\frac{1}{3x^3}\] and so
\[\int_1^{\infty}x^{-4}dx=\lim_{l\to \infty}-\frac{1}{3l^3}+\frac{1}{3}=0+\frac{1}{3}=\frac{1}{3}\]

Answer 15

Yes thanks - got that one okay!
Then used int. by parts for the third which seemed to work