Asked by Vanessa
Evaluate the definite integral.
9x e^(x^2)dx between (0,3)
9x e^(x^2)dx between (0,3)
Answers
Answered by
Kuai
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
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
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