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

Answer 1

Paisley Johnson

#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#

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Answer 2

Christopher Allers

To find the arc length of the curve ( f(x) = x - \sqrt{e^x - 2 \ln x} ) on the interval ([1, 2]), we'll use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} , dx ]

First, we need to find ( f'(x) ), then we'll plug it into the formula and integrate over the interval ([1, 2]).

[ f'(x) = 1 - \frac{1}{2\sqrt{e^x - 2\ln x}} \left(2e^x - \frac{2}{x}\right) ]

[ = 1 - \frac{e^x - \frac{1}{x}}{\sqrt{e^x - 2\ln x}} ]

Now, we'll plug this into the formula for arc length:

[ L = \int_{1}^{2} \sqrt{1 + \left(1 - \frac{e^x - \frac{1}{x}}{\sqrt{e^x - 2\ln x}}\right)^2} , dx ]

This integral represents the arc length of the curve ( f(x) ) on the interval ([1, 2]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.