integration的問題

Slope of the curve at (x , y) is xe-x^2

That is, dy/dx = xe-x^2

y = S xe-x^2dx

y = -1/2 S e-x^2d(-x2)

y = -e-x^2/2 + C

Also, the curve passes through (0 , 1)

So, 1 = -e0/2 + C

C = 3/2

Therefore, the equation of the curve is:

y = -e-x^2/2 + 3/2

2009-05-28 12:08:52 補充：
I used the letter 'S' to replace the integral sign