# Arc Length: Definite and Indefinite Integration

## Main Question or Discussion Point

Several authors state the formula for finding the arc length of a curve defined by $y = f(x)$ from $x=a$ to $x=b$ as:

$$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$

Isn't this notation technically wrong, since the RHS is a definite integral, and the LHS is an indefinite integral (family of antiderivatives)?

I understand that no limits are placed on the integral of $ds$ since $ds$ can be defined in several equivalent ways: y as the independent variable, x as the independent variable, parametrically, or in terms of polar coordinates.

But why can't we write the integral as:

$$\int_{x=a}^{x=b} ds$$

Writing it this way makes it explicit that the variable $x$ is changing from $a$ to $b$.

Alternatively, can't we also write it as:

$$\int_P ds$$

Where P is the path I have defined above.

Last edited:

lavinia
Gold Member
You can not put in integration bounds for ds before you know the integral But you could put the subscript P if you like but that is clear anyway.

The integration bounds x = a to x = b are incorrect because you are integrating with respect to s not with respect to x.

You can not put in integration bounds for ds before you know the integral But you could put the subscript P if you like but that is clear anyway.

The integration bounds x = a to x = b are incorrect because you are integrating with respect to s not with respect to x.
What does carrying out indefinite integration with respect to s mean?

$$\int ds = s + C$$

For $C ∈ ℝ$

What does the right hand side represent?

pasmith
Homework Helper
Several authors state the formula for finding the arc length of a curve defined by $y = f(x)$ from $x=a$ to $x=b$ as:

$$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$

Isn't this notation technically wrong,
Yes. The left hand side should be $$L = \int_0^L\,ds = \int_a^b \frac{ds}{dx}\,dx$$ where $L$ is the length of the curve in question.

• PFuser1232
lavinia
Gold Member
What does carrying out indefinite integration with respect to s mean?

$$\int ds = s + C$$

For $C ∈ ℝ$

What does the right hand side represent?
All I am saying is that the left hand side does not mean an indefinite integral. Perhaps it is an abuse of notation but the integration bounds are implicit. You do not know the integral of ds until you do the integration. You could put it in as an unknown as pasmith did but in practice you can not do this integral with respect to ds. You need a change of parameter where you do know the bounds. But the thing to keep in mind is that an indefinite integral is not meant here.

Yes. The left hand side should be $$L = \int_0^L\,ds = \int_a^b \frac{ds}{dx}\,dx$$ where $L$ is the length of the curve in question.
This makes sense. But if L is the arc length of the curve, what exactly does the variable s represent?

Exactly. Abuse of notation.
This is done over and over again in several texts.

pasmith
Homework Helper
This makes sense. But if L is the arc length of the curve, what exactly does the variable s represent?
If $a \leq x \leq b$ then $s(x)$ is the length of the curve from $(a,f(a))$ to $(x, f(x))$. We have by definition $s(a) = 0$ and $s(b) = L$.

• PFuser1232
If $a \leq x \leq b$ then $s(x)$ is the length of the curve from $(a,f(a))$ to $(x, f(x))$. We have by definition $s(a) = 0$ and $s(b) = L$.
I get it now. It's the same logic used when deriving the area of a circle right?

$$dA = 2πr dr$$
$$\int_0^A dA' = \int_0^r 2πr' dr'$$
$$A = πr^2$$

Stephen Tashi
The variable $s$ must represent distance along the path of the curve, measured from some arbitrary starting pont. For example, there are occasions when you see a curve "parameterized by arc length" in the form (x(s),y(s)).