What is the arclength of $f(x)=x-sqrt(e^x-2lnx)$ on $x in [1,2]$ ?

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Answer 1 · 2017-03-03T04:10:32

$L=int_1^2sqrt(1+(1-(xe^x-2)/(2xsqrt(e^x-2lnx)))^2)dxapprox1.0630$

The arc length of the curve of $f$ on $x in [a,b]$ is given by

$L=int_a^bsqrt(1+(f'(x))^2)dx$

Here, $f(x)=x-(e^x-2lnx)^(1/2)$ so

$f'(x)=1-1/2(e^x-2lnx)^(-1/2)(e^x-2/x)$

$color(white)(f'(x))=1-(xe^x-2)/(2xsqrt(e^x-2lnx))$

Then the arc length is given by

$L=int_1^2sqrt(1+(1-(xe^x-2)/(2xsqrt(e^x-2lnx)))^2)dxapprox1.0630$

What is the arclength of f(x)=x-sqrt(e^x-2lnx)f(x)=x-sqrt(e^x-2lnx) on x in [1,2]x in [1,2]?