Mathematics

# $\displaystyle \int_1^e \log(x) dx= \displaystyle \lim_{n \rightarrow \infty} \displaystyle \sum_{i=1}^n \log \left(1+i\dfrac{e-1}{n} \right)$State whether the above equation is True or False?

False

##### SOLUTION
Given : $\displaystyle \int _{ 1 }^{ e }{ \ln { (x) } dx } = \lim_{ n\rightarrow \infty }\sum _{ i=1 }^{ n }{ \log { \left[ 1+i\cfrac { (e-1) }{ n } \right] } }$
$\displaystyle \int _{ 1 }^{ e }{ \ln { (x) } dx } =\left[ x\ln { x } -x \right] _{ 1 }^{ e }=e-e(D-1)=1\\ \displaystyle \lim_{ n\rightarrow \infty } \sum _{ i=1 }^{ n }{ \log { \left[ 1+i\cfrac { (e-1) }{ n } \right] } }$
$=\displaystyle \lim_{ n\rightarrow \infty }\left[ \log { \left[ 1+i\cfrac { (e-1) }{ n } \right] } +\log { \left( 1+\cfrac { 2(e-1) }{ n } \right) } +\log { \left( 1+\cfrac { 3(e-1) }{ n } \right) } ........... \right]$
$= \displaystyle \lim_{ n\rightarrow \infty } \left[ \cfrac { \log { \left[ 1+i\cfrac { (e-1) }{ n } \right] } \cfrac { (e-1) }{ n } }{ \cfrac { (e-1) }{ n } } +\cfrac { \log { \left( 1+\cfrac { 2(e-1) }{ n } \right) \cfrac { 2(e-1) }{ n } } }{ \cfrac { 2(e-1) }{ n } } +............ \right]$
$=\displaystyle \lim_{ n\rightarrow \infty }\left( 1+2+3+4+........n \right) \left( \cfrac { (e-1) }{ n } \right)$
$=\displaystyle \lim_{ n\rightarrow \infty } \cfrac { n(n+1) }{ 2 } \cfrac { (e-1) }{ n } =\infty$
Hence the statement is false.

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TRUE/FALSE Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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