I'm having trouble with this question on arc length:
y=lnx, (squareroot)3/3 greater than or equal to x less than or equal to 1
It sounds as if you want the length of the y = ln x curve from
x = sqrt(3)/3 (0.57735..) to 1.
The formula for the arc length of a line y(x) is
Length = (INTEGRAL OF) sqrt [1 + (dy/dx)^2] dx
In your case dy/dx = 1/x
so Length = (INTEGRAL OF) sqrt [1 + 1/x^2] dx
Perform the integration between your indicated limits for the answer. I find the integration easier to perform by making the substitution u = 1/x. It's a bit messy.
I have sqrt [1 + 1/x^2] dx , I just don't know what to do next...
Make the substitutions 1/x = u and
dx = u^-2 du
Then the integral becomes
INTEGRAL -[sqrt(1 + u^2)]/u^2 du
According to my table of integrals, the indefinite integral is
[sqrt (1 + u^2)]/u - ln[u + sqrt (u^2 = 1)]
The second term can also be expressed as an arcsinh function.
Integrate that between the two values of u that correspond to your x limits.
Here's a useful site I use to check integrals:
integrals.wolfram. c o m
(delete the spaces)
That's the Wolfram integrator, which is actually Mathematica, a fairly good piece of software for finding anti-derivatives.
When I check sqrt(1 + 1/x^2) I got a slightly different anti-derivative, just thought I'd mention this.
To find the arc length of the curve y = ln(x) from x = sqrt(3)/3 to x = 1, you will need to evaluate the integral of sqrt[1 + (dy/dx)^2] dx, where dy/dx is the derivative of y with respect to x.
In this case, dy/dx = 1/x, so the integral becomes:
∫ sqrt[1 + (1/x)^2] dx
To simplify the integration, you can make the substitution u = 1/x, which implies dx = u^-2 du. Substituting these values into the integral, we get:
∫ sqrt[1 + u^2] u^-2 du
You can now integrate this expression. The indefinite integral of sqrt(1 + u^2) u^-2 du can be found using tables of integrals or computer software like Wolfram Alpha or Mathematica. If you have access to those resources, you can input the expression and get the result.
Alternatively, you can try to solve the integral without these tools. After integrating, you will need to evaluate the result between the values of u that correspond to the x limits given (sqrt(3)/3 and 1). The expression for the indefinite integral is [sqrt(1 + u^2)]/u - ln[u + sqrt(u^2 + 1)]. This second term can also be expressed as an arcsinh function.
To check your calculation, you can use online resources like the Wolfram integrator (integrals.wolfram.com) to verify the result.
It's worth noting that there might be some slight differences in the exact form of the antiderivative due to variations in the calculation methods and software used.