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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*An*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)
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*An*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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here are the ecuations in a easier way to read. \[A=\int\limits_{0}^{1}e^{t^2}dt\] \[B=\int\limits_{0}^{1/2}e^{t^2}dt\] \[I=2\int\limits_{1/2}^{1}[\int\limits_{0}^{x}e^{y^2}dy]dx\] \[I = mA nB +e^{1}e^{1/4}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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 \[f(1/2)=f(1/2)\] \[2f(1)+e^{1}f(1/2)e^{(1/4)}=2AB+e^{1}e^{(1/4)}\] Then m=2 and n=1
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