Why an Initial Value Determines the Constant
Integrating a vector function produces a family of antiderivatives. Each member of that family has the same derivative, because adding a constant vector does not change any component’s derivative. A specified value of the function at one input selects exactly one member of the family.
The calculation is component by component, but the condition is a vector equation: every component of the antiderivative must match the corresponding component of the specified vector at the specified input.
Definitions and the Constant-Vector Form
Let \(\mathbf{f}(t)=\langle f_1(t),f_2(t),\ldots,f_n(t)\rangle\) be a vector-valued function on an interval. An antiderivative of \(\mathbf{f}\) is a vector-valued function \(\mathbf{F}\) satisfying \(\mathbf{F}'(t)=\mathbf{f}(t)\) throughout that interval. Differentiation and integration of vector functions are defined component by component.
The componentwise form of an indefinite integral is therefore
The constants \(C_1,\ldots,C_n\) are independent: the condition on one component does not require its constant to equal the constant in another component. Writing just one scalar \(C\) for the whole vector can wrongly force all components to share the same value.
An initial condition for a vector function specifies both an input \(t_0\) and a vector value \(\mathbf{v}_0\). The condition \(\mathbf{F}(t_0)=\mathbf{v}_0\) means that each component of \(\mathbf{F}(t_0)\) equals the corresponding component of \(\mathbf{v}_0\).
A Reliable Calculation Process
Write a separate constant for each component, or use a constant vector \(\mathbf{C}\).
Substitute \(t=t_0\) into every component of the general antiderivative.
Set corresponding components equal and solve for each constant.
Differentiate the result and substitute \(t_0\) to confirm the derivative and the initial value.
The specified input need not be zero. In particular, do not set the constants equal to the given vector automatically: first evaluate the nonconstant parts of the antiderivative at the specified input.
Worked Examples
Worked Example: Determine a Constant Vector at Zero
Find \(\mathbf{F}(t)\) if \(\mathbf{F}'(t)=\langle 6t-2,\cos t\rangle\) and \(\mathbf{F}(0)=\langle 4,-1\rangle\).
Integrate each component:
Apply the condition at \(t=0\). Since \(3(0)^2-2(0)=0\) and \(\sin 0=0\),
Matching components gives \(C_1=4\) and \(C_2=-1\). Thus
Check the derivative: \(\mathbf{F}'(t)=\langle 6t-2,\cos t\rangle\). Check the condition: \(\mathbf{F}(0)=\langle 4,-1\rangle\). Both requirements hold.
Worked Example: Determine Constants at a Nonzero Input
Find \(\mathbf{G}(x)\) if \(\mathbf{G}'(x)=\langle e^x,4x^3\rangle\) and \(\mathbf{G}(\ln 2)=\langle 5,3\rangle\).
Integrating component by component gives
At \(x=\ln 2\), use \(e^{\ln 2}=2\). The first component condition is \(2+C_1=5\), so \(C_1=3\). The second component condition is \((\ln 2)^4+C_2=3\), so \(C_2=3-(\ln 2)^4\). Therefore,
Differentiate to check: \(\mathbf{G}'(x)=\langle e^x,4x^3\rangle\). Now substitute \(x=\ln 2\): the first component is \(2+3=5\), and the second is \((\ln 2)^4+3-(\ln 2)^4=3\). Hence \(\mathbf{G}(\ln 2)=\langle 5,3\rangle\), as required.
Worked Example: Find Three Independent Constants
Find \(\mathbf{H}(t)\) if \(\mathbf{H}'(t)=\langle 2t,3t^2-4,\cos t\rangle\) and \(\mathbf{H}(2)=\langle 1,0,-3\rangle\).
An antiderivative with all three constants included is
Evaluate each component at \(t=2\): \(2^2+C_1=4+C_1\), \(2^3-4(2)+C_2=8-8+C_2=C_2\), and \(\sin 2+C_3\). Set these equal to the specified components:
Thus \(C_1=-3\), \(C_2=0\), and \(C_3=-3-\sin 2\). The result is
Differentiating gives \(\mathbf{H}'(t)=\langle 2t,3t^2-4,\cos t\rangle\). At \(t=2\), the components are \(4-3=1\), \(8-8=0\), and \(\sin 2-3-\sin 2=-3\). The derivative and the specified value both check.
Why the Initial Value Determines a Unique Antiderivative
The constant-vector rule is more than a convenient way to organize a calculation. On an interval, two antiderivatives of the same vector function can differ only by a constant vector. Here is the justification.
Suppose \(\mathbf{F}'(t)=\mathbf{f}(t)\) and \(\mathbf{G}'(t)=\mathbf{f}(t)\) on an interval \(I\). Define \(\mathbf{D}(t)=\mathbf{F}(t)-\mathbf{G}(t)\). By linearity of differentiation, \(\mathbf{D}'(t)=\mathbf{0}\).
Each component \(D_i(t)=F_i(t)-G_i(t)\) has derivative zero throughout \(I\). For any \(a,b\in I\), the Mean Value Theorem applied between them implies \(D_i(a)=D_i(b)\), so \(D_i\) is constant on \(I\).
Every component of \(\mathbf{D}\) is constant, so \(\mathbf{F}(t)-\mathbf{G}(t)=\mathbf{C}\) for a constant vector \(\mathbf{C}\). Thus antiderivatives differ by a constant vector.
If both antiderivatives satisfy \(\mathbf{F}(t_0)=\mathbf{v}_0\) and \(\mathbf{G}(t_0)=\mathbf{v}_0\), then \(\mathbf{C}=\mathbf{F}(t_0)-\mathbf{G}(t_0)=\mathbf{0}\). Therefore \(\mathbf{F}=\mathbf{G}\) throughout \(I\).
This argument requires the functions to be defined on an interval, so that any two inputs can be joined within the domain. It explains why one vector condition fixes all the constants and leaves only one antiderivative satisfying that condition.
A Common Mistake: Treating the Constant as One Scalar
Suppose an antiderivative has two components. A tempting but incorrect step is to write a single scalar constant in both:
This forces the two independent component constants to be equal. For instance, if the condition is \(\mathbf{F}(0)=\langle 1,5\rangle\), the first component requires \(C=1\), while the second requires \(C=5\). The resulting contradiction is caused by the incorrect shared constant, not by the condition.
The correction is to use two constants:
Then \(\mathbf{F}(0)=\langle C_1,C_2\rangle=\langle 1,5\rangle\), so \(C_1=1\) and \(C_2=5\). The corrected function is \(\langle t^2+1,3t+5\rangle\). Its derivative is \(\langle 2t,3\rangle\), and its value at zero is \(\langle 1,5\rangle\).
Verification Is Part of the Solution
A computed constant vector should be checked in two different ways. Differentiating verifies that the function is an antiderivative of the stated vector function. Substituting the specified input verifies that it satisfies the stated value. Either check alone is incomplete: the right derivative does not guarantee the right initial value, and the right value at one input does not guarantee the right derivative elsewhere.
Check Your Understanding
For each question, show the component equations used to determine the constants. Check your results by differentiating and substituting the given input.
- Find \(\mathbf{P}(t)\) if \(\mathbf{P}'(t)=\langle 4t,2\rangle\) and \(\mathbf{P}(1)=\langle 3,-1\rangle\).
- Why must a vector antiderivative include independent constants for its different components?
- Find \(\mathbf{Q}(x)\) if \(\mathbf{Q}'(x)=\langle \cos x,6x^2\rangle\) and \(\mathbf{Q}(0)=\langle 2,4\rangle\).
- What condition on the domain is used in the uniqueness justification, and where does it enter?
- A student writes \(\langle t^2+C,5t+C\rangle\) as a general antiderivative of \(\langle 2t,5\rangle\). Explain what is wrong and write the correct general form.
