What is the arc length of #f(x)=lnx # in the interval #[1,5]#?

Question

Ezekiel Alexander · Accepted Answer

The arc length is #sqrt26-sqrt2+ln5-ln((1+sqrt26)/(1+sqrt2))# units.

#f(x)=lnx# #f'(x)=1/x# Arc length is given by: #L=int_1^5sqrt(1+1/x^2)dx# Rearrange: #L=int_1^5sqrt(x^2+1)/xdx# Apply the substitution #x=tantheta#: #L=intsectheta/tantheta*sec^2thetad theta# Rewrite as: #L=intcsctheta*(tan^2theta+1)d theta# Hence #L=int(secthetatantheta+csctheta)d theta# Integrate term by term: #L=[sectheta-ln|csctheta+cottheta|]# Reverse the substitution: #L=[sqrt(1+x^2)-ln|(1+sqrt(1+x^2))/x|]_1^5# Insert the limits of integration: #L=sqrt26-sqrt2+ln5-ln((1+sqrt26)/(1+sqrt2))#

Isabella Ackerman · Answer

undefined To find the arc length of the function $ f(x) = \ln(x) $ in the interval [1, 5], we use the formula for arc length:

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

where $ a = 1 $, $ b = 5 $, and $ f'(x) $ is the derivative of $ f(x) $.

The derivative of $ f(x) = \ln(x) $ is $ f'(x) = \frac{1}{x} $.

Substitute the values into the formula:

$$ L = \int_1^5 \sqrt{1 + \left(\frac{1}{x}\right)^2} \, dx $$

$$ L = \int_1^5 \sqrt{1 + \frac{1}{x^2}} \, dx $$

This integral can be solved using trigonometric substitution. Let $ u = \frac{1}{x} $, then $ du = -\frac{1}{x^2} \, dx $.

$$ L = \int \sqrt{1 + u^2} \, (-du) $$

$$ L = -\int \sqrt{1 + u^2} \, du $$

Now we integrate with respect to $ u $:

$$ L = -\frac{1}{2} \left(u \sqrt{1 + u^2} + \ln\left(u + \sqrt{1 + u^2}\right)\right) + C $$

$$ L = -\frac{1}{2} \left(\frac{1}{x} \sqrt{1 + \frac{1}{x^2}} + \ln\left(\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}\right)\right) + C $$

Evaluate this expression from $ x = 1 $ to $ x = 5 $ to find the arc length.