Adding Scalar Integrals Across Piecewise Curves

Why Break a Curve into Pieces?

A curve need not be described by one smooth formula from beginning to end. It may have corners, or it may be convenient to describe different portions using different parameterizations. A polygonal path, for example, is a sequence of line segments joined at vertices. Each segment is smooth, even though the whole path has corners.

For a scalar field \(f\), the scalar line integral accumulates the values of \(f\) along a curve, weighted by distance along the curve. When a curve is made from several smooth pieces, we can calculate the contribution from each piece and add the results. This lets us handle corners without pretending that the whole path is smooth at every point.

Key idea: The scalar line integral over a piecewise smooth curve is the sum of the scalar line integrals over its pieces, in traversal order.

Piecewise Smooth Curves and Scalar Line Integrals

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) describe a curve. The curve is piecewise smooth if there is a finite partition \(a=t_0<t_1<\cdots<t_m=b\) such that \(\mathbf{r}\) is continuously differentiable on each subinterval \([t_{i-1},t_i]\). The derivative may differ from one side of a joining parameter \(t_i\) to the other, so corners are allowed.

Suppose \(f\) is a continuous scalar field on the curve. On a smooth parameterized piece, its scalar line integral is defined by integrating the field value times the speed:

\int_C f\,ds = \int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt.

Here, \(\|\mathbf{r}'(t)\|\) is the speed, and \(ds=\|\mathbf{r}'(t)\|\,dt\) is the arc-length element. For a piecewise smooth curve, the integral is defined by applying this formula on each smooth piece and adding the resulting integrals.

Definition: If the pieces \(C_1,\ldots,C_m\) are traversed in that order, then \(\displaystyle \int_C f\,ds=\sum_{i=1}^{m}\int_{C_i}f\,ds.\) The pieces refer to portions of the traversal, not merely to distinct geometric points: if a path passes through the same point more than once, each occurrence belongs to its corresponding portion.

Why the Additivity Rule Holds

The reason for additivity is the ordinary additivity of definite integrals. On a single parameterized curve, cutting the parameter interval into subintervals cuts the integral into matching parts. Each part uses the same field values and speeds as the corresponding piece of the curve.

Theorem: Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be piecewise smooth with partition \(a=t_0<t_1<\cdots<t_m=b\), and let \(f\) be continuous along its trace. If \(C_i\) is the portion traced for \(t_{i-1}\leq t\leq t_i\), then \(\displaystyle \int_C f\,ds=\sum_{i=1}^{m}\int_{C_i}f\,ds.\)

To justify the result, write the line integral using the parameterization on each smooth interval. On the \(i\)-th interval, the integral is \(\int_{t_{i-1}}^{t_i} f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\). Adding these ordinary definite integrals gives the integral over the full interval, split at the partition points:

\begin{aligned} \sum_{i=1}^{m}\int_{C_i}f\,ds &=\sum_{i=1}^{m}\int_{t_{i-1}}^{t_i} f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\\ &=\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\\ &=\int_C f\,ds. \end{aligned}

The argument does not require the tangent direction to agree across a corner. It only requires the pieces to cover the traversal in order and the line integral to be defined on each piece. The values at the finitely many joining parameters do not contribute a separate term: definite integrals do not acquire extra contributions from isolated endpoints.

Worked Example: Adding the Contributions from Two Pieces

Worked Example: A Two-Piece Path

Let \(f(x,y)=x+2y\). Consider a path that first goes from \((0,0)\) to \((1,0)\), then from \((1,0)\) to \((1,2)\). Parameterize the first piece by \(\mathbf{r}_1(t)=(t,0)\), \(0\leq t\leq1\), and the second by \(\mathbf{r}_2(t)=(1,t)\), \(0\leq t\leq2\).

For the first piece, \(\mathbf{r}_1'(t)=(1,0)\), so its speed is \(\sqrt{1^2+0^2}=1\). Along it, \(f(\mathbf{r}_1(t))=t+2(0)=t\). Therefore,

\int_{C_1} f\,ds =\int_0^1 t(1)\,dt =\left[\frac{t^2}{2}\right]_0^1 =\frac12.

For the second piece, \(\mathbf{r}_2'(t)=(0,1)\), so its speed is \(\sqrt{0^2+1^2}=1\). Along it, \(f(\mathbf{r}_2(t))=1+2t\). Thus,

\int_{C_2} f\,ds =\int_0^2 (1+2t)(1)\,dt =\left[t+t^2\right]_0^2 =(2+4)-0=6.

By additivity, the integral along the whole path is \(\frac12+6=\frac{13}{2}\). As a check, the two parameter intervals cover the two portions in the stated order, and their separate integrals add to the total \(\frac{13}{2}\).

Worked Example: A Continuous Weight on a Polygonal Path

Worked Example: Three Sides of a Polygonal Path

Let \(f(x,y)=1+x+2y\), a continuous scalar field. Let \(C\) travel from \((0,0)\) to \((1,0)\), then to \((1,2)\), and finally to \((3,2)\). We calculate each side’s contribution, then add them.

On the first side, use \(\mathbf{r}_1(t)=(t,0)\), \(0\leq t\leq1\). Its derivative is \((1,0)\) and its speed is \(1\). The field on the side is \(1+t+2(0)=1+t\). Hence

\int_{C_1}f\,ds =\int_0^1(1+t)\,dt =\left[t+\frac{t^2}{2}\right]_0^1 =1+\frac12 =\frac32.

On the second side, use \(\mathbf{r}_2(t)=(1,t)\), \(0\leq t\leq2\). Its derivative is \((0,1)\), so the speed is \(1\). The field becomes \(1+1+2t=2+2t\). Therefore,

\int_{C_2}f\,ds =\int_0^2(2+2t)\,dt =\left[2t+t^2\right]_0^2 =4+4 =8.

On the third side, use \(\mathbf{r}_3(t)=(1+t,2)\), \(0\leq t\leq2\). Its derivative is \((1,0)\), and its speed is \(1\). Along this side, the field is \(1+(1+t)+2(2)=6+t\). Thus

\int_{C_3}f\,ds =\int_0^2(6+t)\,dt =\left[6t+\frac{t^2}{2}\right]_0^2 =12+2 =14.

Adding the three contributions gives

\int_C f\,ds =\frac32+8+14 =\frac{3+16+28}{2} =\frac{47}{2}.

The result is positive, as expected here because \(f=1+x+2y\) is positive along all three sides. The calculation also checks the additivity rule directly: the full path’s integral is exactly the sum of its three piece integrals.

Splitting a Smooth Piece at an Extra Point

A partition point does not have to be a corner. We may split a smooth portion at any parameter value in its interval. The separate values change, but their sum remains the integral over the original portion. This is useful when a calculation becomes simpler after introducing an intermediate point.

Worked Example: Inserting a Partition Point

Consider \(\mathbf{r}(t)=(t,t^2)\), \(0\leq t\leq1\), and the continuous field \(f(x,y)=\sqrt{1+4x^2}\). The derivative is \(\mathbf{r}'(t)=(1,2t)\), so the speed is \(\sqrt{1+4t^2}\). Along the curve, \(x=t\), hence \(f(\mathbf{r}(t))=\sqrt{1+4t^2}\). The product of field value and speed is \(1+4t^2\).

Without an extra split, the integral is

\int_C f\,ds =\int_0^1(1+4t^2)\,dt =\left[t+\frac{4t^3}{3}\right]_0^1 =1+\frac43 =\frac73.

Now insert a partition point at \(t=\frac12\). The first part contributes

\int_0^{1/2}(1+4t^2)\,dt =\left[t+\frac{4t^3}{3}\right]_0^{1/2} =\frac12+\frac{4(1/8)}{3} =\frac12+\frac16 =\frac23.

The second part contributes

\int_{1/2}^1(1+4t^2)\,dt =\frac73-\frac23 =\frac53.

Adding the parts gives \(\frac23+\frac53=\frac73\), matching the unsplit calculation. As a check on the antiderivative, \(\frac{d}{dt}(t+\frac{4t^3}{3})=1+4t^2\), the integrand used on both subintervals.

A Common Mistake: Dropping the Speed

The parameter differential \(dt\) is not automatically the same as distance \(ds\). In the line-integral formula, the speed \(\|\mathbf{r}'(t)\|\) converts parameter change into arc length. Omitting it can make even the contribution of a single piece incorrect.

For a concrete example, let \(f(x,y)=1\) and parameterize the path from \((0,0)\) to \((2,0)\) by \(\mathbf{r}(t)=(2t,0)\), \(0\leq t\leq1\). A wrong calculation might use \(\int_0^1 1\,dt=1\), as if each unit of parameter were one unit of distance. But \(\mathbf{r}'(t)=(2,0)\), so the speed is \(2\), not \(1\). The correct calculation is

\int_C 1\,ds =\int_0^1 1\cdot 2\,dt =2.

This is consistent with the path’s length: it travels two units horizontally. For a piecewise curve, include the speed on each piece before adding the results.

How to Organize a Piecewise-Curve Calculation

Before integrating, make the structure of the traversal explicit. Record each piece’s endpoints, parameter interval, and parameterization. Then evaluate the scalar field and speed on that piece. Add the resulting definite integrals only after each contribution is computed.

List the pieces in traversal order.
Check that the endpoint of each piece is the starting point of the next.

Parameterize each smooth piece.
Use a parameter interval that runs over that portion of the path.

Find the field value and speed.
Compute \(f(\mathbf{r}_i(t))\) and \(\|\mathbf{r}_i'(t)\|\) for each piece.

Integrate and add.
Evaluate each definite integral, then sum the contributions.

Continuity of the scalar field ensures its values are well behaved along the pieces, including at their shared endpoints. A corner does not require an extra term: it is simply where one smooth piece ends and the next begins. The essential bookkeeping is to use the right piece on each interval and retain its speed factor.

Check Your Understanding

For each question, identify the relevant pieces and explain how their contributions combine.

State the additivity rule for a curve divided into four smooth pieces.
Why is a corner permitted in a piecewise smooth curve?
A smooth parameterized curve is split at an interior parameter value. What happens to the sum of its two integrals compared with the original integral?
For \(\mathbf{r}(t)=(3t,0)\), \(0\leq t\leq1\), what speed factor must be used in the scalar line-integral formula?
Why should the pieces be listed in traversal order rather than treated only as a collection of geometric points?