Yes, that' s right. First you should find the Taylor series representation of e^x. Then, in that series, you should replace x with x^2. Lastly, you should integrate it. It is simple, isn't it?Just because [itex]e^{x^2}[/tex] has no antiderivative among the elementary functions does not mean that it has no antiderivative, period. You can solve it as a power series.
ah, but that only works for the complete integral from 0 to ∞, doesn't it?Multiply by [tex]e^{y^2}[/tex], then convert to polar coordinates, and take the sqrt at the end. Thats the usual ansatz for this one.
Well, as D H indicated, it does have an antiderivative, so let's call it g(t).The whole problem states to find the partial derivative of f(x,y) with respect to x,
where, f(x,y)= [tex]\int ^{y}_{x} e^{t^{2}} dt[/tex]
That only works when extending out to infinity or zero. If you try setting actual x and y bounds on the region R, you end up integrating e^(sec[x]^2) and another one just like that. It doesn't simplify, and if you want to integrate from zero to infinity, I'll tell you right now that the answer is infinity. You could expand the integrand as a power series and use termwise integration, however. That might not be terribly helpful in expressing the answer, but it does work.Multiply by [tex]e^{y^2}[/tex], then convert to polar coordinates, and take the sqrt at the end. Thats the usual ansatz for this one.
And if it's e^(-x^2). The integral of e^(x^2) from 0 to infinity is obviously divergent.ah, but that only works for the complete integral from 0 to ∞, doesn't it?
Well, this is an entirely different problem. There is no need to calculate the integral of [itex]\exp(t^2)[/itex] with respect to [itex]t[/itex]. Just use the fundamental theory of calculus.The whole problem states to find the partial derivative of f(x,y) with respect to x,
where, f(x,y)= [tex]\int ^{y}_{x} e^{t^{2}} dt[/tex]
And if it's e^(-x^2). The integral of e^(x^2) from 0 to infinity is obviously divergent.ah, but that only works for the complete integral from 0 to ∞, doesn't it?