# Integration - Dr. Evil is my teacher

1. Nov 9, 2009

### steve2212

1. The problem statement, all variables and given/known data

Right I won't bore you with a sob story but this is a regular grade 12 class and we've been posed with this problem:

Find the arc length of this function across first 5 seconds (t = seconds)

s(t) = t3 - 6t2 + 9t + 5

I learned the formula of s = Integral of (1+f'(x)^2)1/2 * dx

So I began using the Riemann sums until I get here, I'm not quite sure how to proceed.

http://img26.imageshack.us/img26/6931/formulaz.jpg [Broken]

If need be I will show how I got up to there, it just takes forever to use LaTeX.

PS: That should be x->infinity

Thanks!

Last edited by a moderator: May 4, 2017
2. Nov 9, 2009

### latentcorpse

I think it will be easier if you use the formula you were given

Arc Length $:= \int_0^t \left( 1 + s'(t)^2 \right)^{\frac{1}{2}} dt$

now just sub in for $s'(t)$ (which you'll need to work out.

also the upper limit on the integral will become the time that you want to go to. in this case 5 seconds.

3. Nov 9, 2009

### steve2212

Is the integral going to be:

$$\frac{2}{3} (x+3(x^{2}-4x+3)^3)^\frac{3}{2}$$

for the function

$$(1+9(x^{2}-4x+3)^2)^\frac{1}{2}$$

Last edited: Nov 9, 2009
4. Nov 9, 2009

### Staff: Mentor

Your integrand is correct, but should be in terms of t, not x. IOW, it should be (1 + 9(t2 - 4t + 3)2)1/2.

What you show for the antiderivative of the function above is incorrect. You can verify this by taking the derivative of what you showed, and seeing if you get (1 + 9(t2 - 4t + 3)2)1/2.

5. Nov 9, 2009

### steve2212

I tried my best and got this as my final integral, however Mathmetica shows a 14 page integral with complex roots..

$$\frac{2}{3}\left [ x+3 \frac{(x^2-4x+3)^2}{\frac{x^3}{3} - 2x^2 + 3x} \right ]^\frac{3}{2}$$

6. Nov 9, 2009

### Staff: Mentor

Obviously, this is a difficult integral to evaluate exactly. Are you supposed to get the exact answer, or can you use an approximation technique such as Simpson's rule?

Or can you just leave it as an unevaluated definite integral?

7. Nov 10, 2009

### LCKurtz

Perhaps everyone is misinterpreting the problem. Usually a formula such as:

s(t) = t3 - 6t2 + 9t + 5

would describe the position of an object at time t. And instead of "arc length" you would mean the distance traveled. This makes sense for a grade 12 problem. And the catch is that the total distance traveled between t = 0 and t = 5 is not simply s(5) - s(0). That gives the change in position but the object reverses course part of the time. That's the calculus problem -- you have to include the distance going backwards and retracing your position.