Question

Can someone check it I am doing this correctly?

Answer 1

whats the questions

Answer 2

\[\int\limits_{-\infty}^{\infty} 9xe^{-x^2}dx\]

Answer 3

Determine whether the integral is convergent or divergent.
If it is convergent, evaluate it. (If the quantity diverges, enter DIVERGES.)


So I break up the integral
\[\int\limits_{-\infty}^{0} 9xe^{-x^2}dx + \int\limits_{0}^{\infty} 9xe^{-x^2}dx\]

Answer 4

Well here, It was easier to write it down. Did i do everything correctly so far?
and if so.... What do i do next? D:

Answer 5

well first test whether it is convergent or not, if convergent, then the given function is an odd function.

Answer 6

So it is divergent?

Answer 7

Is it necessary to write it as a proper integral before integrating? I can't remember. Because boy.. that sure is a lot of extra writing :\ limit limit limit limit...

Answer 8

it will be 0

Answer 9

Ooo hold on now... how did you end up with x^2 in the denominator?

Answer 10

its an odd function.

Answer 11

Did you integrate e^u into e^u/u for some reason?

Answer 12

Uh... Yea

Answer 13

\[\int\limits_{-a}^{a} f(x) dx =0 if f is odd.. \]

Answer 14

now integrate in a and then let a tends to 0

Answer 15

but you have to test the convergence first

Answer 16

Idk, My professor said when handling infinity to put it into limit notation.

Answer 17

Ok, well anyway, that was the mistake you made.
The last integration step.\[\large\rm \int\limits e^u~du\ne \frac{e^u}{u}\]

Answer 18

Oh isn't\[\int\limits e^u = e^u\]

Answer 19

Yes.

Answer 20

Jango, How do i test for convergence?

Answer 21

Okay i fixed it. But now wut? Eh :/

Answer 22

\[\int\limits_{-a}^{a} xe ^{-x^2} dx =0 \]

Answer 23

for any real value of a. letting a tends to infinity, the above integral is zero and hence the given integral is zero

Answer 24

I am confused...
Because this isn't how my professor taught me how to do it at all...

Answer 25

how it e^infinity = zero? :/

Answer 26

e^- infinity=0

Answer 27

your method is also correct

Answer 28

in your method, from the first part you will get -9/2 and from the second part you will get 9/2

Answer 29

tomorrow I will post a better solution.

Answer 30

Okay

Answer 31

\[f(x)=9x e ^{-x^2}\]
\[f(-x)=-9xe ^{-(x)^2}=-9xe ^{-x^2}=-f(x)\]
hence f(x) is odd function.
\[\int\limits_{-a}^{a}f(x)dx=0\]

Answer 32

@jango_IN_DTOWN Why does e^- infinity=0 and e^infinity=0 ?

Answer 33

\[e ^{-\infty }=\frac{ 1 }{ e ^{\infty} }=\frac{ 1 }{ \infty }\rightarrow 0\]

Answer 34

Oh I see.Thank you, but what about e^inifinty?

Answer 35

\[e ^{^{\infty}}\rightarrow \infty \]

Answer 36

I dont know what I am doing wrong, but i am not getting zero.

Answer 37

\[\int\limits_{-a}^{a} f(x)dx=\int\limits_{-a}^{0}f(x)dx+\int\limits_{0}^{a}f(x)dx\]
f(x) is odd.
so f(-x)=-f(x)
\[I1=\int\limits_{-a}^{0}f(x)dx\]
put x=-t
dx=-dt
when x=-a;t=a
when x=0,t=0
\[I1= \int\limits_{a}^{0}f(-t)(-dt)=-\int\limits_{a}^{0}f(-t)dt=\int\limits_{0}^{a}f(-t)dt=\int\limits_{0}^{a}f(-x)dx=-\int\limits_{0}^{a}f(x)dx\]
so\[\int\limits_{-a}^{a}f(x)dx=-\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(x)dx=0\]

Answer 38

You're making a lot of little mistakes in here.
You should leave the -9/2 out front when you do all of this work.
It seems like all those extra negatives are causing problems.