## anonymous 4 years ago Given A=integrate from 0 to 1 e^(-t^2) dt B=integrate from 0 to 1/2 e^(-t^2) dt compute the following dobule integral: 2 * integrate from -1/2 to 1 [integrate from 0 to x e^(-y^2) dy]dx as a function of A and B. Check that exists positives integers m and n and: I= m*A-n*B+e^(-1)-e^(-1/4) I tried reversing the integral order of the double integral to be able to compute easier, but it gave me: e^(-1)-e^(-1/4) + f(x) + c And if you made in the given order you got: e^(-1)-e^(-1/4) + f(y) + c And f(y) is where A and B came from, so I don't know how to solve this and... (to be continued)

1. anonymous

And I can't compute integrate from 0 to x [e^-(y^2)] dy I think I'll try using the definition of double integral and trying to work with the sumatories, but if anyone can help me I'll be happy. PD: Both constants haven't to be the same, it was a mistake to use "c" in both e^(-1)-e^(-1/4) + f(x) + c e^(-1)-e^(-1/4) + f(y) + k should be better.

2. anonymous

Here are the ecuations in a easier way to read. $A=\int\limits_{0}^{1}e^{-t^2}dt$ $B=\int\limits_{0}^{1/2}e^{-t^2}dt$ $I=2\int\limits_{-1/2}^{1}[\int\limits_{0}^{x}e^{-y^2}dy]dx$ $I = mA -nB +e^{-1}-e^{-1/4}$

3. anonymous

I had an idea. First I define: f(x)=$f(x)=\int\limits_{0}^{x} e^{-y^2}dy$e^(-y^2) I'm evaluating the integral of $e^{-y^2}$ between 0 and x, and this antiderivate in 0 is a constant, so: $f'(x)=e^{-x^2}$ Then I can use integration by parts: $2\int\limits_{}^{}f(x)dx=2(xf(x)-\int\limits_{}^{}xf'(x)dx)$ $2\int\limits_{}^{}xf'(x)dx=2\int\limits_{}^{}xe^{-x^2}dx=-e^{-x^2}$ You can solve the last part with a simple substitution, -x^2=u $2\int\limits_{}^{}xe^{-x^2}dx=2\int\limits\limits\limits_{}^{}xe^{-u^2}/(-2x)du=-\int\limits\limits_{}^{}e^{u}du=-e^u=-e^{-x^2}$ Then: $2\int\limits_{}^{}f(x)dx=2xf(x)-(-e^{-x^2})$ $2\int\limits\limits_{-1/2}^{1}f(x)dx=2(1)f(1)+e^{-1^2}-(2(-1/2)f(-1/2)+e^{-(-1/2)^2})=$ $2f(1)+e^{-1^2}+f(-1/2)-e^{-(1/4)}$ $B=\int\limits_{-1/2}^{0}e^{-t^2}dt \rightarrow -B=\int\limits_{0}^{1/2}e^{-t^2}dt$ Because t is squared, so $e^{-t^2}=e^{-(-t^2)}$ And the function is simetric. Then $f(-1/2)=-f(1/2)$ $2f(1)+e^{-1}-f(1/2)-e^{-(1/4)}=2A-B+e^{-1}-e^{-(1/4)}$ Then m=2 and n=1