Given-I-n-npi-n-1-pi-e-x-sinx-dx-n-N-1-Find-a-relation-between-I-n-1-and-I-n-

Question Number 155496 by mathocean1 last updated on 01/Oct/21

G i v e n I_{n} = \underset{n π}{\int^{(n + 1) π}} e^{- x} s i n x d x, n \in N .

1 . F i n d a r e l a t i o n b e t w e e n I_{n + 1} a n d I_{n} .

Answered by ArielVyny last updated on 01/Oct/21

I_{n} = \int_{n π}^{(n + 1) π} e^{- x} s i n x d x

d u = e^{- x} \to u = - e^{- x}

v = s i n x \to d v = c o s x

I_{n} = {[- e^{- x} s i n x]}_{n π}^{(n + 1) π} + \int_{n π}^{(n + 1)} e^{- x} c o s x d x

d u = e^{- x} \to u = - e^{- x}

v = c o s x \to d v = - s i n x

2 I_{n} = - {[e^{- x} s i n x + e^{- x} c o s x]}_{n π}^{(n + 1) π}

2 I_{n} = - [e^{- π (n + 1)} s i n ((n + 1) π) + e^{- π (n + 1)} c o s ((n + 1) π) - e^{- n π} s i n (n π) - e^{- n π} c o s (n π)]

2 I_{n} = - [e^{- π (n + 1)} (s i n (n π) c o s π + c o s (n π) s i n π) + e^{- π (n + 1)} (c o s n π c o s π + s i n (n π) s i n (π) - e^{- n π} s i n (n π) - e^{- n π} c o s (n π)]

2 I_{n} = - [- e^{- π (n + 1)} {(- 1)}^{n} - e^{- n π} {(- 1)}^{n}]

2 I_{n} = {(- 1)}^{n} e^{- π (n + 1)} + {(- 1)}^{n} e^{- n π}

2 I_{n + 1} = {(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n} e^{- (n + 1) π}

2 I_{n + 1} - 2 I_{n} = {(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n + 1} e^{- n π}

I_{n + 1} - I_{n} = \frac{{(- 1)}^{n + 1} e^{- π (n + 2)} + {(- 1)}^{n + 1} e^{- n π}}{2}

Commented by mathocean1 last updated on 22/Oct/21

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