How do you solve the double integral of xe^(xy) bounded by x=0, y=1, x^2-y=0?

[stanford1463]

How do you solve the double integral of xe^(xy)? bounded by x=0, y=1, x^2-y=0. I used both x-simple and y-simple methods but neither worked..I don't know what the limits are...thanks!

 

[jeffreydk]

You can set up the integral like this for dy dx order...

\int_0^1\int_{x^2}^1xe^{xy} dy dx

Give that a try.

 

[stanford1463]

Cool, thanks! I used dy dx order, but cannot find out what the integral of e^(x^3) is..Is it possible?

 

[jeffreydk]

Well an integral like that is not going to be integrable in terms of elementary functions. I know that something like...

\int e^{-x^2}dx

is an integral that is a gaussian distribution which you can find in a table but that's the closest thing I can think of to the integral you mentioned.

 

[jostpuur]

Sometimes it happens, that writing an iterated integral in some order leads into difficult integrals, while writing it in some other order leads into easy integrals. So one should always simply go through different ways of writing the iterated integral. In this case it seems that integrating over x and y in other order doesn't make it easier though... (if I tried it right) so I don't know what to do there.

One alternative is to leave the number

<br /> \int\limits_0^1 dx\; e^{x^3} \in \mathbb{R}<br />

in the solution like that.

 

[stanford1463]

Well, I did the dy dx order that you mentioned, and the answer came out to be e-1-int (e^(x^3)) (sorry don't know latex). So, I do not know how to solve the integral, from 0 to 1..thanks

 

[stanford1463]

^ that above post was for jeffreydk, and for jostpuur, yes I tried both methods, neither worked out. The original problem wanted the volume under f(x,y)=xe^(xy), over region bounded by x=0, y=1, x^2-y=0...so Idk what is wrong.

 

[jostpuur]

One idea is to write the integral as an integral of 1 over a subset of \mathbb{R}^3.

This is very interesting, because if we get a solution to the original problem, it also gives the precise value of the integral of e^(x^3) over [0,1] in a more explicit form

 

[jeffreydk]

Yea that integral is messy. I tried to get a numerical integration of the entire double integral you wanted and it's complex; I get

\int_0^1\int_{x^2}^1xe^{xy} dy dx = 0.376377+ (3.9255\times 10^{-16})\imath

 

[jostpuur]

Yea that integral is messy. I tried to get a numerical integration of the entire double integral you wanted and it's complex; I get

\int_0^1\int_{x^2}^1xe^{xy} dy dx = 0.376377+ (3.9255\times 10^{-16})\imath

This numerical evaluation doesn't look quite complete You should continue with equation 10^{-16} = 0 next.

 

[jostpuur]

stanford1463, how did you encounter this task? Is this exercise in some book? I have difficulty believing that this would have been put into a book, because that seriously doesn't look like a kind of integral one could actually calculate.

Or well... since the integration domain is bounded, one way to carry out the integration is to use Taylor series of the integrand, but I'm not sure if having the result in form of series, whose value we don't know in more explicit form, would be any better than having it in a form of integral (whose value is not known in more explicit form).

Well, anyway, like this:

<br /> \int\limits_0^1 dx\; e^{x^3} = \sum_{n=0}^{\infty} \frac{1}{n!}\frac{1}{3n+1}<br />

?

 

[stanford1463]

Alright, thanks for the help! Turns out it was a coincidental error by my teacher, who made up this problem.