let I_n = ∫_0 ^n (((−1)^([x]) )/((2x+1)^2 ))dx 1) calculate I_n interms of n 2) find lim_(n→+∞) I


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 37896 by abdo mathsup 649 cc last updated on 19/Jun/18

$l e t I_{n} = \int_{0}^{n} \frac{{(- 1)}^{[x]}}{{(2 x + 1)}^{2}} d x$ $1) c a l c u l a t e I_{n} i n t e r m s o f n$ $2) f i n d {l i m}_{n \to + \infty} I_{n}$
Commented by prof Abdo imad last updated on 19/Jun/18
$1) I_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) (((−1)^k )/((2x+1)^2 ))dx =Σ_(k=0) ^(n−1) (−1)^k [−(1/(2(2x+1)))]_k ^(k+1) =−(1/2) Σ_(k=0) ^(n−1) (−1)^k { (1/(2k+3)) −(1/(2k+1))} =−(1/2)Σ_(k=0) ^(n−1) (((−1)^k )/(2k+3)) +(1/2)Σ_(k=0) ^(n−1) (((−1)^k )/(2k+1)) but Σ_(k=0) ^(n−1) (((−1)^k )/(2k+3)) = (1/3) +Σ_(k=1) ^(n−1) (((−1)^k )/(2k+3)) =_(k−1=p) (1/3) +Σ_(p=0) ^(n−2) (((−1)^(p+1) )/(2p+1)) =(1/3) − Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) ⇒ I_n = −(1/6) +(1/2) Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) +(1/2)Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) +(((−1)^(n−1) )/(2n−1)) I_n =−(1/6) +(((−1)^(n−1) )/(2n−1)) +Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) 2)lim_(n→+∞) I_n =−(1/6) +Σ_(k=0) ^∞ (((−1)^k )/(2k+1)) =(π/4) −(1/6) .$
$1) I_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{{(- 1)}^{k}}{{(2 x + 1)}^{2}} d x$ $= \sum_{k = 0}^{n - 1} {(- 1)}^{k} {[- \frac{1}{2 (2 x + 1)}]}_{k}^{k + 1}$ $= - \frac{1}{2} \sum_{k = 0}^{n - 1} {(- 1)}^{k} {\frac{1}{2 k + 3} - \frac{1}{2 k + 1}}$ $= - \frac{1}{2} \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{2 k + 3} + \frac{1}{2} \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{2 k + 1} b u t$ $\sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k}}{2 k + 3} = \frac{1}{3} + \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{2 k + 3}$ $=_{k - 1 = p} \frac{1}{3} + \sum_{p = 0}^{n - 2} \frac{{(- 1)}^{p + 1}}{2 p + 1}$ $= \frac{1}{3} - \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1} \Rightarrow$ $I_{n} = - \frac{1}{6} + \frac{1}{2} \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1} + \frac{1}{2} \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1} + \frac{{(- 1)}^{n - 1}}{2 n - 1}$ $I_{n} = - \frac{1}{6} + \frac{{(- 1)}^{n - 1}}{2 n - 1} + \sum_{k = 0}^{n - 2} \frac{{(- 1)}^{k}}{2 k + 1}$ $2) {l i m}_{n \to + \infty} I_{n} = - \frac{1}{6} + \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1}$ $= \frac{π}{4} - \frac{1}{6} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$[x] = 0 1 > x ⩾ 0$ $[x] = 1 2 > x ⩾ 1$ $[x] = 2 3 > x ⩾ 2$ $t h u s t h e v a l u e o f {(- 1)}^{[x]} i s e i t h e r + 1 o r - 1$ $\int_{0}^{1} \frac{1}{{(2 x + 1)}^{2}} d x + \int_{1}^{2} \frac{- 1}{{(2 x + 1)}^{2}} + \int_{2}^{3} \frac{1}{{(2 x + 1)}^{2}} + . . .$ $+ \int_{n - 1}^{n} \frac{{(- 1)}^{n - 1}}{{(2 x + 1)}^{2}} d x$ $c o n t d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum