If f(x) is an anti-derivative (i.e. primitive) of x*e^(-x^2) and f(0)=1. Then f(1)=?
let u = x^2
we have
f = 1/2∫e^-u du
f(x) = -1/2 e^(-x^2) + C
1 = -1/2 (1) + C
C = 3/2
f(x) = -1/2 e^(-x^2) + 3/2
f(1) = -1/2 * 1/e + 3/2
= 1/2 (3 - 1/e)
we have
f = 1/2∫e^-u du
f(x) = -1/2 e^(-x^2) + C
1 = -1/2 (1) + C
C = 3/2
f(x) = -1/2 e^(-x^2) + 3/2
f(1) = -1/2 * 1/e + 3/2
= 1/2 (3 - 1/e)