Antiderivatives of Vector-Valued Functions

Why Integrate a Vector-Valued Function?

A vector-valued function can change in several coordinates at once. Finding an antiderivative reverses differentiation in each coordinate: the result is another vector-valued function whose derivative equals the original function.

The central idea is direct. If a function has components \(f_1(t), f_2(t), \ldots, f_n(t)\), find an ordinary scalar antiderivative for each component. Then combine those scalar antiderivatives into a vector. As in scalar calculus, the derivative does not detect added constants; here, the constants form a constant vector.

Definition and Notation

Let \(I\) be an interval of real numbers, and let \(\mathbf{f}:I\to\mathbb{R}^n\) be a vector-valued function. A function \(\mathbf{F}:I\to\mathbb{R}^n\) is an antiderivative of \(\mathbf{f}\) on \(I\) if \(\mathbf{F}\) is differentiable on \(I\) and \(\mathbf{F}'(t)=\mathbf{f}(t)\) for every \(t\in I\).

The vector derivative is computed componentwise. Consequently, the antiderivative problem reduces to \(n\) scalar antiderivative problems. The domain matters: scalar antiderivatives must be defined on the interval under consideration, and any formula used must be valid throughout that interval.

When each component has an antiderivative, the indefinite-integral notation summarizes the resulting family:

where \(\mathbf{F}'(t)=\mathbf{f}(t)\) and \(\mathbf{C}\) is an arbitrary constant vector in \(\mathbb{R}^n\). The notation refers to a family of functions, not to one unique answer. If the component functions are continuous on an interval, the scalar antiderivative results ensure that each has an antiderivative there.

Worked Example: Integrating Polynomial Components

The Componentwise Rule

The preceding calculation gives the general procedure. Suppose each scalar component \(f_i\) has an antiderivative \(F_i\) on the same interval \(I\). Assemble those functions into \(\mathbf{F}(t)=\langle F_1(t),\ldots,F_n(t)\rangle\). The componentwise derivative rule gives \(\mathbf{F}'(t)=\langle F_1'(t),\ldots,F_n'(t)\rangle=\langle f_1(t),\ldots,f_n(t)\rangle=\mathbf{f}(t)\).

Use scalar integration formulas in each coordinate. For example, the power rule applies to \(t^m\) when \(m\ne -1\); the antiderivative of \(\sin t\) is \(-\cos t\); and the antiderivative of \(\cos t\) is \(\sin t\). Each formula is applied to its own component, with the same domain restrictions as in scalar calculus.

Worked Example: Trigonometric Components

Why the Constant Is a Vector

In one-variable scalar calculus, any two antiderivatives of the same function on an interval differ by a constant. The vector version follows by applying that fact separately to each coordinate. Here is the argument in full.

The interval assumption is important in the uniqueness statement. A derivative-zero function is constant on each interval of its domain. If a domain is disconnected, constants on separate pieces need not agree. For the usual antiderivative family on an interval, one arbitrary constant vector describes all possibilities.

Worked Example: Exponential and Reciprocal Components

A Common Mistake: One Constant Instead of a Constant Vector

A frequent error is to append only one scalar constant to the whole vector, as if the same number had to be added to every component. That does not describe the full family. For example, a constant vector can have unrelated entries in its coordinates.

For the constant function \(\mathbf{q}(t)=\langle 1,2\rangle\), a correct general antiderivative is \(\langle t,2t\rangle+\langle C_1,C_2\rangle\). A mistaken restricted answer is \(\langle t,2t\rangle+C\), if this means adding the same scalar to both entries. It omits valid antiderivatives such as \(\langle t+3,2t-5\rangle\).

Differentiate the omitted example to see that it is valid: \(\frac{d}{dt}\langle t+3,2t-5\rangle=\langle 1,2\rangle=\mathbf{q}(t)\). The correct notation \(\mathbf{C}\) allows each component to have its own constant. Differentiating the general answer gives \(\langle 1,2\rangle+\mathbf{0}\), as required.

A Reliable Procedure

For a vector-valued indefinite integral, keep the calculation organized by coordinate.

1. Write down the vector components and the interval on which the function is defined.
2. Find a scalar antiderivative for each component, respecting any domain restrictions.
3. Put those antiderivatives back into vector form and add an arbitrary constant vector.
4. Differentiate the proposed vector componentwise and compare it with the original function.

This method works because vector differentiation is componentwise, and because each scalar antiderivative may be adjusted by its own constant. It does not require finding a special geometric direction or combining the coordinates into a single scalar problem.

Why Antiderivatives Matter

Antiderivatives provide a way to reverse a known vector derivative. At this stage, the essential skill is recognizing that the coordinates can be handled independently while the result remains one vector-valued function. The constant vector records all the freedom left by differentiation.

Always distinguish between an antiderivative, which is one function whose derivative is the given function, and the general antiderivative, which includes every such function on the interval. The derivative check is the simplest reliable way to catch a sign error, a missing chain-rule factor, or an incorrect component.