Question

Integral of e^(-2)?

Answer 1

e^(-x^2) sorry

Answer 2

Is this a definite integral?

Answer 3

yes

Answer 4

What are the limits of the definite integral? I mean what is the value of "a" and "b"
\[\int\limits_{a}^{b}\]

Answer 5

If you have bounds than this integral can be solved, otherwise you can only use an analytic approach through the definition of \(e^x\)

\[\large e^{x}= \sum_{n=0}^{+ \infty} \frac{x^n}{n!} \] with a proper substitution. You might want to share the bounds for a geometric approach.

Answer 6

no a and b

Answer 7

If there is no a and b then it is not definite XD nevermind then. Let's integrate this.

Answer 8

it's e^(x^2)

Answer 9

sorry

Answer 10

What level of Calculus are you on @linyu ?

Answer 11

e to the power of x squared

Answer 12

culc II

Answer 13

In this problem, u = -x^2
Are you with me so far?

Answer 14

okay

Answer 15

so e^u

Answer 16

And by taking the derivative of that, u' = -2x

Answer 17

okay

Answer 18

@KirbyLegs, you won't find a closer expression for this integral.

Answer 19

so i let u=(-x^2)

du/dx = -2x

dx=(1/-2x)du

but x=sqrt(u)

so dx = [1/(-2*sqrt(u))]*du

so integral of e^(-x^2) dx = (e^u)*[1/(-2*sqrt(u))]du

after that i do integration by parts both ways but eventually end up with

(e^u)*[1/(-2*sqrt(u))]du = (e^u)*[1/(-2*sqrt(u))]du

Answer 20

Just let that soak in XD

Answer 21

if \(u=-x^2\)
then how is \(x = \sqrt{u}\) ?
\[\large u=-x^2 \implies x^2=-u \implies x =\sqrt{-u}=i\sqrt{u} \]
but \(x \in \mathbb{R}\) would be a contradiction to that.

Answer 22

http://www.wolframalpha.com/input/?i=int+e%5E%7Bx%5E2%7D
again, there is no closed expression for this integral.

Answer 23

Wow that is very true Spacelimbus . . .
I was kind of ignoring the subtraction sign.

I just did a little research. Linyu's problem is know as the Gaussian integral. You can look at it on Wikipedia
http://en.wikipedia.org/wiki/Gaussian_integral

Answer 24

im lost XD

Answer 25

Does the Gaussian integral or Euler-Poisson integral ring a bell to you? This problem you posted is very special.

Answer 26

That would be true if the integration is closed upon a integrable domain, however, as @Linyu has stated she has no bounds given. So there are a couple of possible scenarios here.

First: Linyu, have you been exposed to Power Series and their rigorous analytic approach to integrals? If not, you cannot solve that integral, simple as that.

Secondly: You did not copy the problem correctly, you might have an additional factor of \(x\) in front of the integral, or you have given goths

However, in all cases, you need a lot of Analysis and definitely more than just basic integration skills.

Answer 27

sorry, the auto correct of OpenStudy somehow turned my boths into goths :D

Answer 28

According to wikipedia, the bounds are from -infinity to +infinity. Those would be the "a" and "b" values

Answer 29

so it must be a definite integral?

Answer 30

Correct. Or you could have a +C at the end of the equation, representing many possible functions. It is easier and more "definite" with limits.

Answer 31

If it is a definite integral, then there is a geometric approach to solve that problem, but if and only if the limits are plus and minus infinity. This is rather a Calc 3 Problem than a Calc 2 Problem however.

If you have no bounds given, you need to leave the geometric interpretation and switch into an analytic interpretation with power series and study their convergence behavior first (completeness of \(\mathbb{R}\), completeness of the \(\exp\) function and uniform convergence).

Answer 32

Lol i'll let you take this one from here Spacelimbus. I'm only in Calc 1 :)

Answer 33

it's actually a bonus question

Answer 34

it just isn't a beginners problem, and I don't mean to cause any mayhem on here, but the way it is stated above it is sold as "easy to do", it is not, in no approach whatsoever.

Answer 35

If the problem is
\[ I= \int_{-\infty}^{\infty} e^{-x^2} dx\]
we can do this trick:
\[ I^2= \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dy \\
= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)} dx \ dy
\]
in polar coordinates, this is
\[ \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r\ dr \ d\theta \]

Answer 36

that can be integrated, and gives the value for I^2. take the square root for I

Answer 37

@phi can you list of chapter I should look up for this question?

Answer 38

I do not know what you are studying. But here is a short video that shows how to integrate exp(-x^2)

http://ia700409.us.archive.org/33/items/MIT18_02SCF10/MIT18_02SCF10Rec_37_300k.mp4