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### Course: Integral Calculus>Unit 3

Lesson 13: Arc length

# Worked example: arc length

A finely tuned example demonstrating how the arc length formula works.

## Want to join the conversation?

• Why was the bounds of the integral changed on ? Would keeping the original bounds still work when you replace u back after taking the integral of u with respect to the original bounds?
• The original bounds were in terms of x. Once we did the u-sub, we are now in terms of u, so to find the new bounds in terms of u we did the process you referenced.
In answer to your second question, YES, if you want to convert back to terms of x after you do the integration in terms of u, then you can use the original bounds.
Do what works best for you.
• Why did Sal not revert from the u-sub back to the original values in terms of x? The problem began in terms of x, it should end in terms of x also I feel.
• Good question.
You'll notice that around Sal changes the interval of the integral into values of u.
This part was sort of "optional". He could just as well have left the integral boundaries in terms of x and back substituted at the end of the problem. The important thing is that this has to be consistent or you will get the wrong answer.

Also this situation is specific to the definite integral. The answer is just a number so there is no x or u value in the final answer. If this were an indefinite integral and you left the answer in terms of u then that would be pretty much wrong. For those problems you should back substitute and leave the answer in terms of x.
• I wanted to play around with this method for calculating the arc length of a simple y=x^2 parabola and chose the boundaries of 0 and 2...
So first step, you know the derivative of x^2 is 2x and you have to square that derivative in the formula, so you get 4x^2.
Plug in the interval and that derivative squared, and you have the integral from 0 to 2 of √(4x^2+1).
After this, I've tried integration by parts and some u substitution, maybe I missed learning how to solve integrals like this, but I don't know how to solve the integral from 0 to 2 of √(4x^2+1). I'd really love some help!
• This turns out to be a pretty tricky integral. It can be done via trig substitution, but you need to know how to find the integral of the secant function and what to do when the original integral reappears after using integration by parts .

There's a nice step-by-step explanation of how to find the integral of sqrt(1+x^2), which is basically the same problem, here:

• how would you find the arc length, when you have two variables for example if you have x=t^3, and y=4t ?
• The arclength in the x-y plane is ALWAYS ∫ √( dx² + dy²). Thus, if you are given x(t) and y(t) (we say "parametric" equations for x and y), then we can write this as:
L = ∫ √( (dx/dt)² + (dy/dt)² ) dt
Basically, we have "divided" everything inside the radical by dt², and so we then multiply on the outside of the radical simply by dt. Since x and y are functions of t, this will be a purely "t" integral.
• Does the arc length formula s = rθ come from applying this arc length formula to circles?

P. S. At , Sal draws a gorgeous integral sign :)
• Yes, it does! If you use this formula on f(x) = sqrt(r^2 - x^2) over a certain interval of the curve, you will get just this result. Very interesting how everything connects together. :)
• If the equation had been y^2=x^3 and then you were asked to find the integral from point a to b. Would you have to consider both curves? Since when you take the square root you would have y=x^3/2 and y=-x^3/2.

So would we have to consider both curves? Thus, doubling our answer for the top portion since they are symmetric?
• It would depend on your definition of "point A to point B" and on your variable of integration. Knowing that y^2=x^3 is not a one-to-one function in terms of y=f(x), you would need to decide whether you were looking for the length of both halves of the curve or just one between two given x-values. (That is, if you're integrating with respect to x.) If you were trying to find the length of the curve between two specific points (x,y), then yes, you would need to account for both parts of the curve.

If you're integrating with respect to y, on the other hand, your equation would give you the length of both parts of the curve without any extra trouble (given a lower and upper limit for y) because this can be expressed as a one-to-one function in terms of x=f(y). So in this case, integration with respect to y would probably be a better choice for representing the true length of the curve between any two given points.

Hope that helps.
• why does the sqrt of u = u^3/2?
(1 vote)
• The square root of u is u^1/2. The derivative of u^3/2 is (3/2)u^1/2, so the anti-derivative of u^1/2 is (u^3/2)/(3/2) or (2/3)u^3/2.
• At , shouldn't it be the absolute value of x? Squaring the square root should produce an absolute value right? Or at least restrict the domain to non-negative x-values.
(1 vote)
• In this case, x is positive, so abs(x) = x.
• I was thinking about this integrals when we derive int ds we get a int b sqrt(1+(f'(x))^2)dx isn't dx suppose to multiply rather than being left out like that? Or calculus is made for almost an obvious answers?
(1 vote)
• When you are writing and rearranging an integral, you want to get a differential (such as dx) multiplied out and placed at the end of the integral. This is the standard form of an integral, as it shows what you are integrating with respect to, and then you can start to find the value of the integral.
Hope this helps!