Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 97307 by eidmarie last updated on 07/Jun/20

Commented by john santu last updated on 07/Jun/20

$\underset{1}{\int^{\infty}} (\frac{\ln (5 x + n) - \ln (n)}{n}) dx$
Commented by bemath last updated on 07/Jun/20

${[x \ln (5 x + 1) - x + \frac{1}{5} \ln (5 x + 1)]}_{0}^{1}$ $(\ln (6) - 1 + \frac{1}{5} \ln (6)) - 0$ $\frac{6}{5} \ln (6) - 1$
	Answered by Sourav mridha last updated on 07/Jun/20

	$\lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} l n [\frac{5 k}{n} + 1] = \int_{0}^{1} l n (5 x + 1) d x$
	Commented by ahmedeid last updated on 07/Jun/20

	$c a n y o u [s o l v e [i t [b y [s t e p s [s i r$
	Commented by bemath last updated on 07/Jun/20

	$using D . I method$
	Commented by Sourav mridha last updated on 07/Jun/20
	$∫_0 ^1 ln(5x+1)dx=(1/5)∫_0 ^1 ln(5x+1)d(5x+1) = =(1/5)[(5x+1){ln(5x+1)−1}]_0 ^1 =(1/5)[6(ln6−1)+1]$
	$\int_{0}^{1} l n (5 x + 1) d x = \frac{1}{5} \int_{0}^{1} l n (5 x + 1) d (5 x + 1)$ $= = \frac{1}{5} {[(5 x + 1) {l n (5 x + 1) - 1}]}_{0}^{1}$ $= \frac{1}{5} [6 (l n 6 - 1) + 1]$
	Commented by ahmedeid last updated on 07/Jun/20

	$t h a n k s [a l o t [s i r$
	Commented by ahmedeid last updated on 07/Jun/20

	Commented by ahmedeid last updated on 07/Jun/20

	$n u m b d r 14 p l e a s e$
	Commented by bemath last updated on 07/Jun/20

	$qn 97305$
	Commented by Sourav mridha last updated on 07/Jun/20
	welcome
	Answered by mathmax by abdo last updated on 08/Jun/20
	$let S_n =Σ_(k=1) ^n (1/n)(ln(5k+n)−ln(n)) ⇒ S_n =(1/n) Σ_(k=1) ^n ln(((5k+n)/n)) =(1/n) Σ_(k=1) ^n ln(1+5(k/n))→∫_0 ^1 ln(1+5x)dx lim_(n→+∞) S_n =∫_0 ^1 ln(1+5x)dx changement ln(1+5x)=t give 1+5x =e^t ⇒x =((e^t −1)/5) ⇒∫_0 ^1 ln(1+5x)dx =∫_0 ^(ln6) t ((1/5))e^t dt =(1/5) ∫_0 ^(ln6) t e^t dt =(1/5){ [t e^t ]_0 ^(ln6) −∫_0 ^(ln6) e^t dt} =(1/5){6ln6 −[e^t ]_0 ^(ln6) } =(6/5)ln6 −(1/5)(6−1) =(6/5)ln6 −1$
	$let S_{n} = \sum_{k = 1}^{n} \frac{1}{n} (\ln (5 k + n) - \ln (n)) \Rightarrow$ $S_{n} = \frac{1}{n} \sum_{k = 1}^{n} \ln (\frac{5 k + n}{n}) = \frac{1}{n} \sum_{k = 1}^{n} \ln (1 + 5 \frac{k}{n}) \to \int_{0}^{1} \ln (1 + 5 x) dx$ $\lim_{n \to + \infty} S_{n} = \int_{0}^{1} \ln (1 + 5 x) dx changement \ln (1 + 5 x) = t give$ $1 + 5 x = e^{t} \Rightarrow x = \frac{e^{t} - 1}{5} \Rightarrow \int_{0}^{1} \ln (1 + 5 x) dx = \int_{0}^{\ln 6} t (\frac{1}{5}) e^{t} dt$ $= \frac{1}{5} \int_{0}^{\ln 6} t e^{t} dt = \frac{1}{5} {{[t e^{t}]}_{0}^{\ln 6} - \int_{0}^{\ln 6} e^{t} dt}$ $= \frac{1}{5} {6 \ln 6 - {[e^{t}]}_{0}^{\ln 6}} = \frac{6}{5} \ln 6 - \frac{1}{5} (6 - 1) = \frac{6}{5} \ln 6 - 1$
	Commented by bemath last updated on 08/Jun/20

	$yes . we answer the same$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum