let-I-n-0-1-x-n-e-x-dx-with-n-integr-natural-1-calculate-I-0-I-1-and-I-2-2-find-arelation-between-I-n-and-I-n-3-find-I-n-interms-of-n-

Question Number 66330 by mathmax by abdo last updated on 12/Aug/19

l e t I_{n} = \int_{0}^{1} x^{n} e^{- x} d x w i t h n i n t e g r n a t u r a l

1) c a l c u l a t e I_{0}, I_{1} a n d I_{2}

2) f i n d a r e l a t i o n b e t w e e n I_{n} a n d I_{n}

3) f i n d I_{n} i n t e r m s o f n .

Commented by mathmax by abdo last updated on 18/Aug/19

1) I_{n} = \int_{0}^{1} x^{n} e^{- x} d x \Rightarrow I_{0} = \int_{0}^{1} e^{- x} d x = {[- e^{- x}]}_{0}^{1} = 1 - e^{- 1}

I_{1} = \int_{0}^{1} x e^{- x} d x =_{b y p a r t s} {[- {x e}^{- x}]}_{0}^{1} + \int_{0}^{1} e^{- x} d x = - e^{- 1} + 1 - e^{- 1}

= 1 - 2 e^{- 1}

I_{2} = \int_{0}^{1} x^{2} e^{- x} d x =_{b y p s r t s} {[- x^{2} e^{- x}]}_{0}^{1} + \int_{0}^{1} 2 x e^{- x} d x

= - e^{- 1} + 2 (1 - 2 e^{- 1}) = 2 - 3 e^{- 1}

2) b y p a r t s u = x^{n} a n d v^{^{'}} = e^{- x} \Rightarrow I_{n} = {[- x^{n} e^{- x}]}_{0}^{1} + \int_{0}^{1} {n x}^{n - 1} e^{- x} d x

= - e^{- 1} + n \int_{0}^{1} x^{n - 1} e^{- x} d x = n I_{n - 1} - \frac{1}{e} \Rightarrow I_{n} = {n I}_{n - 1} - \frac{1}{e}

3) l e t V_{n} = \frac{I_{n}}{n!} w e h a v e V_{n + 1} - V_{n} = \frac{I_{n + 1}}{(n + 1)!} - \frac{I_{n}}{n!}

= \frac{(n + 1) I_{n} - \frac{1}{e}}{(n + 1)!} - \frac{I_{n}}{n!} = \frac{I_{n}}{n!} - \frac{1}{e (n + 1)!} - \frac{I_{n}}{n!} = - \frac{1}{e (n + 1)!} \Rightarrow

\sum_{k = 0}^{n - 1} (V_{k + 1} - V_{k}) = - \frac{1}{e} \sum_{k = 0}^{n - 1} \frac{1}{(k + 1)!} = - \frac{1}{e} \sum_{k = 1}^{n} \frac{1}{k!} \Rightarrow

V_{n} - V_{0} = - \frac{1}{e} \sum_{k = 1}^{n} \frac{1}{k!} b u t V_{0} = I_{0} = 1 - \frac{1}{e} \Rightarrow

V_{n} = - \frac{1}{e} (\sum_{k = 1}^{n} \frac{1}{k!} + 1) + 1 = 1 - \frac{1}{e} \sum_{k = 0}^{n} \frac{1}{k!} \Rightarrow

I_{n} = n! {1 - \frac{1}{e} \sum_{k = 0}^{n} \frac{1}{k!}} (n ⩾ 1)