Evaluate the integral: int_0^1 (x^2 +1)e^(-x)dx

Alex Baird 2022-07-28 Answered
Evaluate the integral:
0 1 ( x 2 + 1 ) e x d x =
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Answers (1)

Ali Harper
Answered 2022-07-29 Author has 16 answers
0 1 ( x 2 + 1 ) e x d x = 0 1 x 2 e x d x + 0 1 e x d x
0 1 e x d x = ( e x ) 0 1 = e 1 + 1
0 1 x 2 e x d x = [ u = x 2 u = 2 x , v = e x v = e x ] = u v u v = x 2 e x + 2 x e x d x
2 x e x d x = [ u = 2 x u = 2 , v = e x v = e x ] = 2 x e x + 2 e x d x
0 1 x 2 e x d x = ( x 2 e x ) 0 1 ( 2 x e x ) 0 1 + 0 1 2 e x d x = e 1 2 e 1 + 2 ( e 1 + 1 ) = 5 e 1 + 1
0 1 ( x 2 + 1 ) e x d x = 0 1 x 2 e x d x + 0 1 e x d x = 5 e 1 + 1 e 1 + 1 = 6 e 1 + 2
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