Linearity and Additivity of Vector Integrals

Why These Rules Matter

Definite vector integrals combine component integrals into a single vector. Their algebraic rules let us reorganize a calculation before evaluating it: constants can be factored out, sums can be separated, and an interval can be divided into simpler pieces. Reversing the endpoints changes the direction of accumulation and negates the result.

These rules are useful only when applied with the right conditions and signs. In particular, reversing endpoints changes the sign of the whole vector, not just one of its components.

Definition and Conditions

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be continuous, and write its components as \(r_1,\ldots,r_n\). Its definite integral is defined by integrating each component over the same interval:

\int_a^b \mathbf{r}(t)\,dt = \left( \int_a^b r_1(t)\,dt,\, \int_a^b r_2(t)\,dt,\, \ldots,\, \int_a^b r_n(t)\,dt \right).

Continuity on the interval is a sufficient condition for these component integrals to exist. The rules below apply when the vector functions involved are integrable on the relevant intervals; continuity is a convenient condition that ensures this in the examples here. When an endpoint is split into two intervals, the functions must be integrable on both pieces as well.

Definition: A definite vector integral is computed componentwise. Equality between two vector integrals therefore means that their corresponding components are equal.

Linearity: Constants and Sums

Let \(\mathbf{u}\) and \(\mathbf{v}\) be integrable vector functions on \([a,b]\), and let \(\lambda\) be a constant scalar. The constant-multiple and sum rules are

\int_a^b \lambda\mathbf{u}(t)\,dt = \lambda\int_a^b \mathbf{u}(t)\,dt, \qquad \int_a^b \bigl(\mathbf{u}(t)+\mathbf{v}(t)\bigr)\,dt = \int_a^b \mathbf{u}(t)\,dt + \int_a^b \mathbf{v}(t)\,dt.

More generally, for constants \(\alpha\) and \(\beta\),

\int_a^b \bigl(\alpha\mathbf{u}(t)+\beta\mathbf{v}(t)\bigr)\,dt = \alpha\int_a^b \mathbf{u}(t)\,dt + \beta\int_a^b \mathbf{v}(t)\,dt.

These statements follow from the corresponding scalar integral rules, applied separately to each component. The constants must not depend on \(t\): a scalar function of \(t\) inside the integrand cannot in general be taken outside the integral.

Worked Example: Integrating a Linear Combination

Let \(\mathbf{F}(t)=(t,\,2t-1)\) and \(\mathbf{G}(t)=(3,\,t^2)\). Compute \(\int_0^2 \bigl(2\mathbf{F}(t)-\mathbf{G}(t)\bigr)\,dt\), using linearity and then checking by direct componentwise integration.

First form the vector integrand:

2\mathbf{F}(t)-\mathbf{G}(t) = (2t,\,4t-2)-(3,\,t^2) = (2t-3,\,4t-2-t^2).

Using the constant-multiple and sum rules gives

2\int_0^2 \mathbf{F}(t)\,dt-\int_0^2 \mathbf{G}(t)\,dt.

Compute the two integrals componentwise:

\int_0^2\mathbf{F}(t)\,dt = \left(\left[\frac{t^2}{2}\right]_0^2,\,[t^2-t]_0^2\right) = (2,\,2),

\int_0^2\mathbf{G}(t)\,dt = \left([3t]_0^2,\,\left[\frac{t^3}{3}\right]_0^2\right) = \left(6,\,\frac{8}{3}\right).

Therefore,

2(2,\,2)-\left(6,\,\frac{8}{3}\right) = (4,\,4)-\left(6,\,\frac{8}{3}\right) = \left(-2,\,\frac{4}{3}\right).

To check, integrate the combined integrand directly:

\int_0^2(2t-3,\,4t-2-t^2)\,dt = \left([t^2-3t]_0^2,\, \left[2t^2-2t-\frac{t^3}{3}\right]_0^2\right) = \left(-2,\,\frac{4}{3}\right).

The direct calculation agrees with the result from linearity.

Additivity: Splitting an Interval

Suppose \(a\leq c\leq b\), and \(\mathbf{r}\) is integrable on \([a,b]\). Additivity across adjacent intervals states that

\int_a^b \mathbf{r}(t)\,dt = \int_a^c \mathbf{r}(t)\,dt + \int_c^b \mathbf{r}(t)\,dt.

The point \(c\) can be chosen to make the computation easier, but it must lie between the endpoints. The integral on the whole interval is the sum of the contributions from the two pieces. If \(c=a\) or \(c=b\), one of the pieces has zero length and contributes zero.

Worked Example: Splitting an Integral at an Interior Point

Let \(\mathbf{r}(t)=(t^2,\,2t,\,1-t)\). Split its integral from \(0\) to \(3\) at \(1\), compute both pieces, and verify that their sum equals the integral over the full interval.

On the first interval,

\int_0^1\mathbf{r}(t)\,dt = \left( \left[\frac{t^3}{3}\right]_0^1,\, [t^2]_0^1,\, \left[t-\frac{t^2}{2}\right]_0^1 \right) = \left(\frac{1}{3},\,1,\,\frac{1}{2}\right).

On the second interval,

\int_1^3\mathbf{r}(t)\,dt = \left( \left[\frac{t^3}{3}\right]_1^3,\, [t^2]_1^3,\, \left[t-\frac{t^2}{2}\right]_1^3 \right) = \left(\frac{26}{3},\,8,\,-2\right).

Adding the two vectors gives

\left(\frac{1}{3},\,1,\,\frac{1}{2}\right) + \left(\frac{26}{3},\,8,\,-2\right) = \left(9,\,9,\,-\frac{3}{2}\right).

For a check, integrate directly from \(0\) to \(3\):

\int_0^3\mathbf{r}(t)\,dt = \left( \left[\frac{t^3}{3}\right]_0^3,\, [t^2]_0^3,\, \left[t-\frac{t^2}{2}\right]_0^3 \right) = \left(9,\,9,\,-\frac{3}{2}\right).

The direct integral equals the sum of the two pieces, as additivity requires.

Reversing the Endpoints

For an integrable vector function, the integral with reversed endpoints is defined by changing the sign:

\int_b^a \mathbf{r}(t)\,dt = -\int_a^b \mathbf{r}(t)\,dt.

This convention also gives \(\int_a^a\mathbf{r}(t)\,dt=\mathbf{0}\), where \(\mathbf{0}\) is the zero vector. Reversal changes the sign of every component because the scalar integral in every component changes sign.

Worked Example: Reversing the Limits

Let \(\mathbf{r}(t)=(2t-1,\,t^2)\). Compute \(\int_3^1\mathbf{r}(t)\,dt\), and verify the result both by reversing the forward integral and by evaluating directly.

First compute the forward integral:

\int_1^3\mathbf{r}(t)\,dt = \left( [t^2-t]_1^3,\, \left[\frac{t^3}{3}\right]_1^3 \right) = \left(6,\,\frac{26}{3}\right).

Reversing the limits negates the entire vector:

\int_3^1\mathbf{r}(t)\,dt = -\left(6,\,\frac{26}{3}\right) = \left(-6,\,-\frac{26}{3}\right).

To verify this by direct evaluation with the reversed limits,

\int_3^1\mathbf{r}(t)\,dt = \left( [t^2-t]_3^1,\, \left[\frac{t^3}{3}\right]_3^1 \right) = \left((0)-6,\,\frac{1}{3}-9\right) = \left(-6,\,-\frac{26}{3}\right).

Both methods give the same result.

Why the Rules Work

The identities are consequences of scalar integration. The following derivation proves linearity; the same componentwise reasoning gives additivity and endpoint reversal.

Write each vector by components.
Let \(\mathbf{u}(t)=(u_1(t),\ldots,u_n(t))\) and \(\mathbf{v}(t)=(v_1(t),\ldots,v_n(t))\). Their sum has components \(u_i(t)+v_i(t)\), for \(i=1,\ldots,n\).

Apply scalar linearity to each component.
For each \(i\), \(\int_a^b(u_i(t)+v_i(t))\,dt=\int_a^b u_i(t)\,dt+\int_a^b v_i(t)\,dt\).

Reassemble the components.
The component equalities together state exactly that \(\int_a^b(\mathbf{u}(t)+\mathbf{v}(t))\,dt=\int_a^b\mathbf{u}(t)\,dt+\int_a^b\mathbf{v}(t)\,dt\).

For additivity, scalar additivity gives \(\int_a^b r_i(t)\,dt=\int_a^c r_i(t)\,dt+\int_c^b r_i(t)\,dt\) for each component \(i\). Combining those equalities gives the vector identity. For endpoint reversal, scalar reversal gives \(\int_b^a r_i(t)\,dt=-\int_a^b r_i(t)\,dt\) for every \(i\); therefore, the vector integral also changes sign. Thus the vector rules hold because each component obeys the corresponding scalar rule.

Theorem: If the vector functions involved are integrable on the stated intervals, definite vector integration is linear, additive across adjacent intervals, and changes sign when its endpoints are reversed. Each identity follows by applying the corresponding scalar integral identity to every component.

A Common Sign Mistake

A frequent error is to reverse the limits but change the sign of only one component. For instance, if \(\int_1^3\mathbf{r}(t)\,dt=(6,\,26/3)\), it would be incorrect to claim that \(\int_3^1\mathbf{r}(t)\,dt=(-6,\,26/3)\). The second component has not been reversed.

The correction is to negate the whole vector:

-\left(6,\,\frac{26}{3}\right) = \left(-6,\,-\frac{26}{3}\right).

A reliable check is to evaluate every component with the reversed endpoints or to place one minus sign in front of the entire forward-integral vector before distributing it.

Using the Rules Together

The rules can be combined, provided their conditions are respected. For example, a linear combination can be integrated on each of two adjacent intervals, and those results can then be added. Reversing the endpoints of the combined integral negates every component of the final vector. Organizing a calculation this way can reduce repeated work, but it does not change what the vector integral means: each component must still give the same value as its scalar integral.

Key takeaway: Constant multiples and sums can be handled componentwise using linearity; adjacent intervals can be joined using additivity; and reversing endpoints negates the entire integral vector.

Check Your Understanding

Answer each question by using the componentwise rules for definite vector integrals.

State the constant-multiple rule for \(\int_a^b \lambda\mathbf{u}(t)\,dt\), and say what \(\lambda\) must be.
If \(a\leq c\leq b\), how can \(\int_a^b\mathbf{r}(t)\,dt\) be written using integrals over \([a,c]\) and \([c,b]\)?
If \(\int_2^5\mathbf{q}(t)\,dt=(3,\,-4,\,1)\), what is \(\int_5^2\mathbf{q}(t)\,dt\)?
Why do vector integral linearity rules follow from scalar integral rules?
What is the value of \(\int_a^a\mathbf{r}(t)\,dt\)?