The area between the curves y -log e x and y -log e x 2 isA- 3-eB- e-log e 3C- e-2D- 1

The correct option is A 3 e nbsp; nbsp; nbsp; Point of intersection: log_e x=(log_ex)^2 ⇒ nbsp;log_e x(log_e x 1)=0 ⇒log_e x=0, log_e x=1 ⇒ x=1, x=e ...

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Byju's Answer

Standard XII

Mathematics

Finding Roots of Quadratic

The area betw...

Question

The area between the curves $y = {log}_{e} x$ and $y = ({log}_{e} x)^{2}$ is

3 - e

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e - {log}_{e} 3

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e - 2

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1

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Solution

The correct option is A $3 - e$

Point of intersection:
${log}_{e} x = ({log}_{e} x)^{2} \Rightarrow {log}_{e} x ({log}_{e} x - 1) = 0 \Rightarrow {log}_{e} x = 0, {log}_{e} x = 1 \Rightarrow x = 1, x = e$

$Area = e \int 1 {log}_{e} x - ({log}_{e} x)^{2} d x$
$= [x ({log}_{e} x - 1) - x ({log}_{e} x)^{2} + 2 x ({log}_{e} x) - 2 x]_{1}^{e} = 3 - e$

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The area between the curves y=logex and y=(logex)2 is

The correct option is A 3−e Point of intersection: logex=(logex)2⇒logex(logex−1)=0⇒logex=0, logex=1⇒x=1, x=e Area=e∫1logex−(logex)2dx =[x(logex−1)−x(logex)2+2x(logex)−2x]e1=3−e

The area between the curves $y = {log}_{e} x$ and $y = ({log}_{e} x)^{2}$ is