Write it as the repeated integral:
Integral over y from 0 to 1 dy Integral over x from 0 to y dx e^(y^2) =
Integral over y from 0 to 1 dy ye^(y^2)
= 1/2 [exp(1) - 1]
Calculate the double Integral
e^(y^2) dA
D = {(x,y)| 0 <= x <= y <= 1}.
9 answers
So I tried to work it out.
(1,0) (0,y) e^y^2 dxdy
=(0,1) x(e^y^2) |(0,y) dy
= (0,1) ye^y^2 dy
u = y^2
du = 2ydy
1/2du = ydy
1/2 (0,1) e^u du
Once I get to this point I don't know what to do.
I put u back in but don't know how to get to the final answer.
1/2 (0,1)(e^y^2)(2ydy)
Is this right?
(1,0) (0,y) e^y^2 dxdy
=(0,1) x(e^y^2) |(0,y) dy
= (0,1) ye^y^2 dy
u = y^2
du = 2ydy
1/2du = ydy
1/2 (0,1) e^u du
Once I get to this point I don't know what to do.
I put u back in but don't know how to get to the final answer.
1/2 (0,1)(e^y^2)(2ydy)
Is this right?
You can integrate ye^y^2 dy by using that the derivative of
e^(y^2) is 2y e^(y^2)
e^(y^2) is 2y e^(y^2)
1/2 (0,1) e^u du
but that is 1/2 e^u evaluated at u = 1 minus 1/2 of e^u evaluated at 0
=1/2(e) - 1/2 (1)
=1/2(e-1)
like Count Iblis said
I sure am glad the Count came along because I was messing with those limits for about 15 minutes :)
but that is 1/2 e^u evaluated at u = 1 minus 1/2 of e^u evaluated at 0
=1/2(e) - 1/2 (1)
=1/2(e-1)
like Count Iblis said
I sure am glad the Count came along because I was messing with those limits for about 15 minutes :)
so if I find the derivative of ye^y^2 do i not need to find u and du?
ok so I do 1/2(e) - 1/2 (1) = 1/2 (e-1)
so would that be my answer? I thought I still had to put the u back in. How did u = 1 minus 1/2 of e^u evaluated at 0? Not sure where that come from
so would that be my answer? I thought I still had to put the u back in. How did u = 1 minus 1/2 of e^u evaluated at 0? Not sure where that come from
The antiderivative of y exp(y^2) equals 1/2 exp(y^2)
ok so the final answer is 1/2 e^y^2????
e^(y^2)