Question

Find the arc length of the curve over the given interval. $$ y=\ln x, \quad[1,5] $$

Short answer

The arc length of the given function over the interval \([1,5]\) is \(\sinh^{-1}(5) - \sinh^{-1}(1).\)

Step-by-step solution

Find the Derivative of the Function

The given function is \(y = \ln x\). Its derivative, obtained using the rule \(d/dx \ln x = 1/x\), is therefore, \(y' = 1/x\).

Substitute the Derivative in the Formula and Evaluate the Integral

Now, we need to substitute \(y'\) into the arc length formula. So, we get \(L = \int_1^5\sqrt{1+(1/x)^2}dx.\) This integral may not be straightforward to solve, a good way would be to simplify the integrand as follows: \[L =\int_1^5 \frac{\sqrt{x^2+1}}{x} dx.\] To solve this punctual integral without exhaustive steps, recognize the integrand as the derivative of arcsinh(x) - i.e, the hyperbolic arcsine of x. Then, L becomes a difference of arcsinh(x) over the interval.

Compute the Arc Length Using the Integral Limits

In this final step, let's use the fundamental theorem of calculus which states that, if a function F(x), is the integral of f(x), over the domain [a,b], then ∫ from a to b of f(x) dx equals F(b) - F(a). Because the integral from step 2 is [arcsinh(x)] from 1 to 5, we can simply substitute these limits to get, \[L = \sinh^{-1}(5) - \sinh^{-1}(1)\].