Question

Hi!
I'm trying to evaluate the double int(1-(y^7)*(e^(-x^5))). I have the boundary of y and x. My problem is with the exp(-x^5). I really don't know what to do with it...

Answer 1

Something tells me that it is impossible to find...

Answer 2

We should be able to integrate any continuous function...

Answer 3

Yes. But it is not 100% that you can get an expression in elementary functions for it. May be you have bounds for an integral? Can you write a full promlem?

Answer 4

\[\int\limits_{0}^{2}\int\limits_{y^2}^{4}(1-y^2e ^{-x^{5}}dxdy)\]

Answer 5

Hm.. Where did you find this?

Answer 6

School xD I am asqued to ''evaluate'' it. I have been looking all over my books and tried several methods...

Answer 7

http://www.wolframalpha.com/input/?i=integrate+e^ {-x^5}
It is too complicated. May be there is a mistake.

Answer 8

Yeah I already tried Wolframalpha xD It wasnt very helpful. I'm sure there is no error in the equation. Thx for the help anyway

Answer 9

Do this first\[\int\limits_{x=y^2}^{x=4}(1-y^2e^{-x^5})dx=\int\limits_{x=y^2}^{x=4}1dx-y^2\int\limits_{x=y^2}^{x=4}e^{-x^5}dx\] The right hand side one is hellish to do- Substitute u=x^5 and integrate by parts?

Answer 10

http://answers.yahoo.com/question/index?qid=20080229185246AAZADzq
Doesn't look good

Answer 11

Hmm interesting. ''The problem you're trying to solve is impossible -- e^(x^5) has no elementary antiderivative. This doesn't mean that it can't be integrated (_any_ continuous function can be integrated)...''

Answer 12

Mathematica out
\[ \text{ConditionalExpression}\left[2 \left(4-\frac{1}{5} y^4 \text{ExpIntegralE}\left[\frac{4}{5},y^{10}\right]+\frac{1}{5} y^2 \left(-5+\text{Gamma}\left[\frac{1}{5},\\1024\right]\right)\right),\left(\frac{y^2}{4-y^2}\notin \text{Reals}\left\|\text{Re}\left[\frac{y^2}{4-y^2}\right]\leq -1\right\|\\\left(\text{Re}\left[\frac{y^2}{4-y^2}\right]\geq 0\&\&\frac{y}{-4+y^2}\neq 0\right)\right)\&\&\\(\text{Re}[y]>2\|\text{Re}[y]<-2\|((\text{Re}[y]==-2\|\text{Re}[y]==2)\&\&y\notin \text{Reals})\|-2\\<\text{Re}[y]<0\|0<\text{Re}[y]<2)\right] \]