Question

can anyone talk with me about improper integrals? i have a question or two and it should only be a few mins

Answer 1

When you are given an integral how do know when to test if it converges or diverges?'

Answer 2

when to test? usually when you are asked: does this converge or diverge :/

Answer 3

ok so i have some questons where its the integral from 0 to infinity. i know to add the limit part when there is an infinity there. But when it is not that obvious when do iadd it? i think vertical asymptotes?

Answer 4

an integral is more compact than discrete stuff; so if the integral version goes to zero; the discrete does too

Answer 5

if i had a problem to work on it be easier to demonstarte

Answer 6

me no type gooder this nite lol

Answer 7

ok so for example integral from 0 to infinity of dx/(SQRT(x)(x+4))

Answer 8

so i took that and because there is an infinity a add the lim as b approces inf of the integral from 0 to b

Answer 9

\[\int_{0}^{inf}\ \frac{1}{(x+4)\sqrt{x}}dx\]

Answer 10

this perchance?

Answer 11

correct

Answer 12

thats bad at zero as well

Answer 13

see thats what i thought but in the key i got from the professor he only added the one that approaches infinity

Answer 14

the infinity part just makes the bottom huge and this thing goes to 0; but for propercalities ...

Answer 15

-4 and 0 are bad

Answer 16

so we need to split this up into prolly 4 integrals to test out

Answer 17

ahhh ok good so im not crazy lol

Answer 18

you dont need to do the work for it its alright i just wanted to make sure i was not under the wrong impression

Answer 19

\[\int_{t}^{1}\frac{1}{(x+4)\sqrt{x}}dx+\int_{1}^{u}\frac{1}{(x+4)\sqrt{x}}dx+\int_{u}^{10}\frac{1}{(x+4)\sqrt{x}}dx +\int_{10}^{w}\frac{1}{(x+4)\sqrt{x}}dx\]

Answer 20

i figure 1 and 10 were good numbers; just need something to measure form so the values themselves just need to be inside the good interval parts

Answer 21

lol... helps if i read; 0 to inf so -4 aint even in the scope

Answer 22

ok i see that makes sense. So when im solving any integral problem is the first thing i should check is if it has vertical asymptotes? because if it does i need to do the limit parts right

Answer 23

lol

Answer 24

right; and in this case -4 is a VA but its not even in our interval to begin with so 0 and inf are the trouble spots

Answer 25

ok makes sense so im just trying to figure out the thought process for integrals

Answer 26

find a good spot to measure from; then limit it out to the trouble spots

Answer 27

what it was always before was Integration by parts, Partial deriv, trig sub or u-sub then solve. but now that we have learned this the first step should be to find vertical asymptotes and split it up then figure out what method to use for integration?

Answer 28

yep, sounds planish to me

Answer 29

ok cool. actually, looking back does it matter? if you find the antideriv and split up by vertical asymptotes before you do the interval you would get the same answer it looks like

Answer 30

the asymps just help to define your plan of attack; so get comfortable with it in any manner you want

Answer 31

since the same curve is being analysed throught out; then yeah, you can integrate first and get it set up for the limit processes

Answer 32

wow i guess i just needed to write it all out to realize it lol. TY forhelping and...listening? lol and making sure i dident say anyting dumb

Answer 33

yw :) and good luck

Answer 34

u too