Question

improper integral problem:

Answer 1

\[\int\limits_{-\infty}^{-\infty}1/(3\sqrt{x})\]

Answer 2

I spose we might have to break this apart into 2 improps?

Answer 3

So far I integrate to get: \[2\sqrt{x}/3 \int\limits_{-\infty}^{-\infty}\]

Answer 4

is this really from -inf to -inf?

Answer 5

Yea

Answer 6

any integral from a to a = 0 right?

Answer 7

So i am not sure how I would evaluate since its negative going into a square root.

Answer 8

Hmm, not sure

Answer 9

\[\int_{5}^{5}e^{x^2}dx = 0 \]

Answer 10

there is no such thing as
\[\int_{-\infty}^{-\infty}\]

Answer 11

\[\int_{124}^{124}x^2+sin(3x^2+4)dx=0\]

Answer 12

yes but infinity is not a number

Answer 13

What I am supposed to do is determine whether the problem "converges" or "diverges"

Answer 14

well, as "b" moves to infinity the limit stuff implies that its 0 or something right?

Answer 15

i mean maybe you are supposed to say 0, but it doesn't make any sense. an improper integral is a limit

Answer 16

I don't know thats what I need to find out

Answer 17

whether the limit approaches infinity (diverges) or a number (converges)

Answer 18

well i don't know where this question comes from, but if you are really supposed to give an answer just say 0 and don't give it another minutes thought. but if you asked me i would say it does not exist

Answer 19

\[\int_{b}^{b}f(x)dx=0\]
as b appraoches infinity, the integral = 0 still right?

Answer 20

yea

Answer 21

So suppose the bottom bound was 0 instead of -infinity

Answer 22

Then I would have to do the methodology I was thinking of, since we couldn't just say 0 by that theorem.

Answer 23

then youd be evaluating a different integral :)

Answer 24

So how could I determine the value limit if when plugging in negative infinity yields an imaginary numer?

Answer 25

\[\int_{b}^{0}f(x)dx=N\]
\[\int_{0}^{b}f(x)dx=-N\]
add them together to get 0 id assume

Answer 26

i havent delved into imaginaries yet :)

Answer 27

Ah, ok

Answer 28

I am thinking mayeb l'hopital's rule

Answer 29

perhaps, but I cant be certain of it .... yet