Question

ganeshie8

Answer 1

this is how u should tag
@UCLA104

Answer 2

first type this @<-- then user name
@username so then they get notification like u just did
@nnesha @UCLA104 <--like this :)

Answer 3

@ganeshie8

Answer 4

^ :)

Answer 5

ohhhhhhhhhh ok thank you I didnt know that

Answer 6

yeah np :) :)
best of luck!!

Answer 7

you're trying to test convergence ?

Answer 8

Yes. Using the comparison test

Answer 9

Would you agree that \(x^3 \gt x+x^2\) for all \(x \ge 2\) ?

Answer 10

Yes

Answer 11

that yields
\[ \dfrac{1}{x^3} \lt \dfrac{1}{x+x^2}\]

yes ?

Answer 12

does the 1/3 exponent not matter?

Answer 13

we will see..

Answer 14

Ok.

Answer 15

which is same as saying :

\[\dfrac{1}{x+x^2}\gt \dfrac{1}{x^3}\]

Answer 16

take cube root both sides now

Answer 17

we get
\[\dfrac{1}{(x+x^2)^{1/3}}\gt \dfrac{1}{x}\]

Answer 18

look up harmonic series and appeal to comparison test

Answer 19

Yes. and because 1/x diverges then the original one diverges?

Answer 20

Thats it!
http://gyazo.com/f0f4103a181b01e4cce91c086a168049

Answer 21

Ok i guess what im confused on is how did you know to use x^3?

Answer 22

thats a good question, yeah thats the only key part here. the reasoning might go something like this :

In the bottom we have \( (x+x^2)^{1/3}\) which we know has degree \(2/3 \lt 1\). So by p-series test this has to diverge. however degree for a radical expression makes no sense, so we're not allowed to use p-series test here. Next best thing we can do is to compare it with a polynomial... thinking of anything other than x^3 is just ridankulous as the radical 1/3 is beggins us to use x^3 for comparison...

Answer 23

Ok. What I was taught is to find which part on the denominator is the dominating factor and to essentially pull it out when using the comparison test

Answer 24

so for the next question, 74, we have 1/x(e^x + x)

Answer 25

after you factor out an x of course

Answer 26

and e^x + x is larger than x

Answer 27

Hey, I remember doing this stuff, and just for reference, this might come in handy.

Answer 28

that looks like a good start

Answer 29

wait give me a sec to send a picture of what i did

Answer 30

disregard the squared xe^x stuff

Answer 31

I think you're absolutely right if the lower bound of integral is \(1\)

but the lower bound is \(0\) here and what do you know about the convergence of \(\int\limits_0^{\infty} \frac{dx}{x^2}\) ?

Answer 32

im desperately hoping that it was a typo lol, if it was really \(0\), then we need to do something else to show that it diverges...

Answer 33

nah its zero lol

Answer 34

then we're screwed.. we need to start over :/

Answer 35

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Answer 36

dos that mean the answer for the first question is wrong too because its also 0 to infinity

Answer 37

good question, for first problem, the logic is slightly off but it shouldnt matter because we have showed that\(\int_1^{\infty}\dfrac{dx}{(x+x^2)^(1/3)}\) diverges... so that also means \(\int_0^{\infty}\dfrac{dx}{(x+x^2)^(1/3)}\) diverges too

so im hoping we are mostly good with first problem...

Answer 38

so does it really make a huge difference in the second problem that its 0 and not 1?

Answer 39

would it help by splitting the integrals apart at all? by making 2 separate integrals from 0 to 1 and 1 to infinity?

Answer 40

It does diverge

Answer 41

\(\int\limits_{\color{REd}{0}}^{\infty} \dfrac{dx}{xe^x+x^2}\) diverges

\(\int\limits_{\color{REd}{1}}^{\infty} \dfrac{dx}{xe^x+x^2}\) converges

Answer 42

the lower bound makes that much difference ^

Answer 43

well damn

Answer 44

so we gotta prove that it diverges

Answer 45

and im really not sure how if what i did doesnt apply to the lower bound being zero :/

Answer 46

Notice the integral is improper at both ends : 0 and infinity

Answer 47

thats exactly the reason why we need to work the lower bound separately

Answer 48

so we split it into 2 integrals?

Answer 49

thats a good idea

Answer 50

then we could take the limit as R goes to zero from the right of the integral r to 1 of the function

Answer 51

and then the other one would be limit as r goes to infinity of the integral 1 to r of the function

Answer 52

\[\int\limits_0^{\infty} \dfrac{dx}{xe^x+x^2} = \int\limits_0^{a}
\dfrac{dx}{xe^x+x^2} + \int\limits_a^{\infty} \dfrac{dx}{xe^x+x^2} \]

Answer 53

How do you do that? like make it all nice and easy to read without having to type it....

Answer 54

yes that will do

Answer 55

but when its in two separate integrals, both of them have to converge or diverge right? because if one diverged when the other converged, it wouldnt work

Answer 56

but that doesnt work because we know when its 1 to infinity it converges

Answer 57

Yeah I am stuck at the same point. Do we have anything like below :

sum of a diverging series and an absolutely converging series is divergent ?

Answer 58

because the integrand is positive in the given bounds

Answer 59

i found a link that says the sum of a convergent and divergent is divergent

Answer 60

which makes sense because it is the same as a constant plus a divergent

Answer 61

Nice :) then we're done

Answer 62

just show that \(\large \int\limits_0^a \dfrac{dx}{xe^x+x^2} \) diverges and \(\int\limits_a^{\infty} \dfrac{dx}{xe^x+x^2}\) converges

Answer 63

how would i show that it diverges exactly?

Answer 64

we can use the same comparison test

Answer 65

exactly but would it converge using the same comparison test?

Answer 66

Im sorry, we just started doing this stuff and im not really sure about it

Answer 67

fix \(a = 1\), we show that \(\large \int\limits_0^a \dfrac{dx}{xe^x+x^2}\) is divergent :

when \(x\in (0,1)\) we have \(\large e^x+x \lt 4 \implies x(e^x+x) \lt 4x \implies \dfrac{1}{x(e^x+x)} \gt \dfrac{1}{4x} \)

Answer 68

ok thank you

Answer 69

did u get why that works

Answer 70

yeah i understand why it works i just wouldnt have thought of it myself

Answer 71

it doesnt matter what number you use :

when \(x\in (0,1)\) :
\(e^x + x \lt 100\) also works but it is easy to see that \(e^x+x \lt 4\) if we recall \(e^1 \lt 3\)

Answer 72

Thank you

Answer 73

poor @username
@Nnesha i'm sure he gets tagged in all of these explanation posts...
;P