Why Integrate a Vector-Valued Function?
A vector-valued function can describe a quantity with several components that change together. For example, a function \(\mathbf{r}(t)\) might give a point in the plane or in space for each input \(t\). If we know the vector-valued function \(\mathbf{v}(t)\), we can ask for a function whose derivative is \(\mathbf{v}(t)\).
The method is direct: find an antiderivative of each component. The important detail is that each component has its own arbitrary constant. Together, those constants form a constant vector, so the indefinite integral represents a family of vector-valued functions rather than just one function.
Definition and Notation
Suppose \(\mathbf{r}\) is a vector-valued function defined on an interval \(I\), with \(n\) component functions:
A vector-valued function \(\mathbf{F}\) is an antiderivative of \(\mathbf{r}\) on \(I\) if \(\mathbf{F}'(t)=\mathbf{r}(t)\) for every \(t\) in \(I\). Differentiation of a vector-valued function is component by component, so this condition means that \(F_i'(t)=r_i(t)\) for every component \(i\).
Equivalently, the integral can be written by placing an integral sign on each component:
Each component integral on the right includes its own arbitrary constant. The vector notation \(\mathbf{C}\) collects these constants in one place.
Worked Example: Integrating Polynomial Components
Worked Example: Integrating Polynomial Components
Find the indefinite integral of \(\mathbf{r}(t)=\langle 3t^2-4,\,2t+5\rangle\), and verify the result.
Integrate the first component. An antiderivative of \(3t^2\) is \(t^3\), and an antiderivative of \(-4\) is \(-4t\). Include a constant \(C_1\):
For the second component, an antiderivative of \(2t\) is \(t^2\), and an antiderivative of \(5\) is \(5t\). Include a separate constant \(C_2\):
Therefore, the vector antiderivative is
Check by differentiating each component:
The derivative of each constant is zero, and the original vector-valued function is recovered.
Why the Constants Form a Vector
For a scalar function, antiderivatives differ by a constant on an interval. For a vector-valued function, this fact applies separately to every component. Here is a direct justification for the component-by-component rule and for the independent constants.
If \(\mathbf{F}(t)=\langle F_1(t),\ldots,F_n(t)\rangle\), then \(\mathbf{F}'(t)=\langle F_1'(t),\ldots,F_n'(t)\rangle\). Thus, if \(F_i'=r_i\) for each \(i\), then \(\mathbf{F}'=\mathbf{r}\).
For any constants \(C_1,\ldots,C_n\), the derivative of \(F_i+C_i\) is \(F_i'\), since the derivative of a constant is zero. Therefore adding \(\mathbf{C}=\langle C_1,\ldots,C_n\rangle\) does not change the derivative.
If \(\mathbf{G}'=\mathbf{r}\) and \(\mathbf{F}'=\mathbf{r}\), then \((G_i-F_i)'=0\) for every component. A differentiable scalar function with derivative zero on an interval is constant there, so \(G_i-F_i=C_i\). Hence \(\mathbf{G}=\mathbf{F}+\mathbf{C}\), with a possibly different constant for each component.
This establishes that componentwise integration gives the full family of antiderivatives on an interval, provided each component has an antiderivative there. It also explains why the constants are independent: the derivative condition places no restriction tying one component's constant to another's.
Worked Example: Trigonometric and Exponential Components
Worked Example: Trigonometric and Exponential Components
Find an indefinite integral of \(\mathbf{r}(t)=\langle 2\cos(3t),\,3\sin t,\,4e^{2t}\rangle\), and verify it.
For the first component, \(\frac{d}{dt}\sin(3t)=3\cos(3t)\). Therefore an antiderivative of \(2\cos(3t)\) is \(\frac{2}{3}\sin(3t)\):
For the second component, \(\frac{d}{dt}(-3\cos t)=3\sin t\), so
For the third component, \(\frac{d}{dt}e^{2t}=2e^{2t}\). Thus an antiderivative of \(4e^{2t}\) is \(2e^{2t}\):
Combining the components gives
Check by differentiating:
Each factor from the chain rule is included: the \(3\) from \(\sin(3t)\) cancels with the denominator \(\frac{2}{3}\), and the \(2\) from \(e^{2t}\) turns \(2e^{2t}\) into \(4e^{2t}\).
Choosing Antiderivatives on an Interval
The antiderivative rules for scalar functions apply to the components, but the domain still matters. A proposed antiderivative must be defined and differentiable on the interval under consideration. For instance, \(\arctan t\) is defined for every real \(t\), and its derivative is \(1/(1+t^2)\), so it can be used as an antiderivative of that component on \(\mathbb{R}\).
As with scalar integration, not every function has an elementary antiderivative. The componentwise definition does not guarantee that every component integral can be expressed using elementary functions. It says that whenever each component has an antiderivative on the interval, assembling those component antiderivatives gives a vector antiderivative.
Worked Example: A Polynomial Component and an Inverse-Tangent Component
Worked Example: A Polynomial Component and an Inverse-Tangent Component
Find the indefinite integral of \(\mathbf{r}(t)=\langle (t+1)^2,\,1/(1+t^2)\rangle\) on \(\mathbb{R}\), and verify it.
For the first component, the power rule gives an antiderivative \((t+1)^3/3\), because differentiating this expression gives \((t+1)^2\). For the second component, \(\frac{d}{dt}\arctan t=1/(1+t^2)\). Attach independent constants:
Now differentiate the first component:
Differentiate the second component:
Consequently, differentiating the vector gives \(\langle (t+1)^2,\,1/(1+t^2)\rangle=\mathbf{r}(t)\), as required.
Common Mistake: Using One Scalar Constant
A frequent error is to use the same scalar constant in every component, as though it represented every possible antiderivative. Consider \(\mathbf{r}(t)=\langle 2t,3\rangle\). Integrating the components gives \(t^2\) and \(3t\), but the general answer is
A tempting but incomplete answer is \(\langle t^2+C,\ 3t+C\rangle\). Differentiating this gives \(\langle 2t,3\rangle\), so it is indeed an antiderivative. However, it describes only the cases where the two component constants are equal. For example, the valid antiderivative \(\langle t^2+4,3t-2\rangle\) cannot be written in that form with one shared \(C\).
Why This Method Matters
Vector-valued functions package several scalar functions into one object. The componentwise method preserves that structure while letting you use familiar scalar antiderivative rules. It also makes verification systematic: differentiate each component and compare it with the corresponding component of the original function.
When writing a solution, keep the components in a consistent order, integrate each one correctly, and include the constant vector. Then check the result by differentiation. That final check catches errors such as a missing chain-rule factor, an incorrect sign, or a component copied into the wrong position.
Check Your Understanding
For each question, use componentwise integration and remember that the constant in each component is independent.
- Find an indefinite integral of \(\langle 6t,\,4\rangle\).
- Find an indefinite integral of \(\langle \sin t,\,2\cos(2t)\rangle\), and verify it by differentiation.
- Explain why \(\langle t^2+C,\,3t+C\rangle\) does not describe every antiderivative of \(\langle 2t,\,3\rangle\).
- What condition must each component antiderivative satisfy for the resulting vector function to be an antiderivative of \(\mathbf{r}(t)\)?
