# Wow I am stumped! What is the difference between these two integrals?

1. Aug 16, 2011

### flyingpig

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

Suppose s'(t) is a velocity function, then which of the integral will give you the total distance?

(1) $$\int_{a}^{b} \sqrt{1 + [s'(t)]^2} dt$$

(2) $$\int_{a}^{b} |s(t)| dt$$

3. The attempt at a solution

No clue at all...

the first is arc length, so it is like summing up the distances into one. The second one is absolute value of area of total displacement. They both are??

2. Aug 16, 2011

### Quinzio

What ?

Neither of them gives you a distance.....
The first one is the lenght of a curve $f(x)$. If it happens that $f(x)$ is a speed, you don't get the distance !!!
Let's take $f(x) = 0$, speed zero.
$$\int_a^{b} \sqrt{1+{|f(x)|}^2}\ dt= \int_a^{b} \sqrt{1+0}\ dt = b-a$$ !!!

Your speed is zero... but you travel some distance (b-a) ???

The second one, you basically multiply a lenght by a time.... and the you integer (you sum). What do you get ??? For sure you don't get a distance....
Under some circumstances you get an area....but... mmmm....
What was it for ?

3. Aug 16, 2011

### Harrisonized

(1) is definitely correct for ℝ². However, if (2) is a typo and you actually meant

ab |s'(t)| dt,

then this is by far a better answer than (1), and I'll show you why.

Let s(t) be a parametric equation defined as (x(t),y(t)). Then the arc length integral is:

ab ds = ∫ab |s'(t)| dt

(Note that ds = |s'(t)| dt)

= ∫ab √( [dx/dt]² + [dy/dt]² ) dt

= ∫ab √( [dx]² + [dy]² )

= ∫ab √( [1]² + [dy/dx]² ) dx

= ∫ab √( 1 + [dy/dx]² ) dx

There isn't a difference in ℝ². However, in ℝn, you generally want to use:

ab |s'(t)| dt = ∫ab √( [dx1/dt]² +... + [dxn/dt]² ) dt

Btw Quinzio, your formula is just plain wrong.

Last edited: Aug 16, 2011
4. Aug 16, 2011

### SammyS

Staff Emeritus
If s'(t) is the velocity function in one dimension, then:
The total distance traveled in the time interval from t1 to t2 is $\displaystyle \int_{t_1}^{\,t_2}\,|s'(t)|\,dt\,.$

The displacement during the same interval is $\displaystyle \int_{t_1}^{\,t_2}\,s'(t)\,dt=s(t_2)-s(t_1)\,.$