Question

Evaluate the definite integral.
9x e^(x^2)dx between (0,3)

Answer 1

u = x^2 Between (0,9)
du = 2x
du/2 = x
[9/2 e^u du ] from 0 t 9
[9/2 e^u ] from 0 t 9
9/2[e^9 -e^0]
9/2[e^9-1]= 36459.3776741