let-A-n-0-n-1-x-2-x-1-dx-1-calculate-A-n-2-find-lim-n-A-n-

Question Number 38804 by maxmathsup by imad last updated on 30/Jun/18

l e t A_{n} = \int_{0}^{n} \frac{{(- 1)}^{x}}{2 [x] + 1} d x

1) c a l c u l a t e A_{n}

2) f i n d {l i m}_{n \to + \infty} A_{n}

Commented by math khazana by abdo last updated on 30/Jun/18

A_{n} = \int_{0}^{n} \frac{{(- 1)}^{[x]}}{2 [x] + 1} d x

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18

\int_{0}^{1} \frac{{(- 1)}^{0}}{2 \times 0 + 2} d x + \int_{1}^{2} \frac{{(- 1)}^{1}}{2 \times 1 + 1} + \int_{2}^{3} \frac{{(- 1)}^{2}}{2 \times 2} + \dots +

Commented by math khazana by abdo last updated on 01/Jul/18

A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{{(- 1)}^{k}}{2 k + 1} d x = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{2 k + 1}

2) {l i m}_{n \to + \infty} A_{n} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} = \frac{π}{4} .

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18

\int_{0}^{1} \frac{{(- 1)}^{0}}{2 \times 0 + 1} d x + \int_{1}^{2} \frac{{(- 1)}^{1}}{2 \times 1 + 1} d x + \int_{2}^{3} \frac{{(- 1)}^{2}}{2 \times 2 + 1} d x + \dots

+ \int_{n - 1}^{n} \frac{{(- 1)}^{n}}{2 (n - 1) + 1} d x

= \frac{1}{1} - \frac{1}{3} + \frac{1}{5} + \dots + \frac{{(- 1)}^{n}}{2 n - 1}

2) w e k n o w {t a n}^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \dots

s o w h e n n \to \infty A_{n} \to {t a n}^{- 1} (1)

s o a n s i s

n \to \infty A_{n} \to \frac{Π}{4}