Why a Vector Integral Is Built from Scalar Integrals
A vector-valued function can have several components, each changing with the input. Integrating the function means accumulating the change in every component. The result is a vector, and the scalar Fundamental Theorem of Calculus applies to each component.
This componentwise viewpoint gives two useful results. First, the derivative of an accumulated vector integral recovers its integrand. Second, if we know a vector antiderivative, we can evaluate a definite vector integral by subtracting its values at the endpoints.
As for scalar integrals, reversing the endpoints reverses the sign: \(\int_a^b\mathbf r(t)\,dt=-\int_b^a\mathbf r(t)\,dt\), and \(\int_a^a\mathbf r(t)\,dt=\mathbf 0\). These statements hold component by component.
The Fundamental Theorem: Variable Upper Endpoint
Suppose \(\mathbf r\) is continuous on an open interval \(I\), and choose a fixed number \(a\in I\). For \(t\in I\), define the accumulated vector integral \(\mathbf R(t)=\int_a^t\mathbf r(u)\,du\). The open interval condition ensures that every \(t\in I\) has nearby inputs on both sides, so the ordinary two-sided derivative is defined there.
The proof is the scalar Fundamental Theorem applied to every component. To see why it works, write \(\mathbf r(t)=\langle r_1(t),\ldots,r_n(t)\rangle\). The \(j\)th component of \(\mathbf R\) is \(R_j(t)=\int_a^t r_j(u)\,du\). For sufficiently small nonzero \(h\), both \(t\) and \(t+h\) are in \(I\), and the difference quotient for this component is
Subtract \(r_j(t)\) from both sides of the difference quotient. By linearity of the scalar integral, its difference from \(r_j(t)\) is the average of \(r_j(u)-r_j(t)\) over the interval from \(t\) to \(t+h\). Continuity makes this average approach zero: given \(\varepsilon>0\), choose \(\delta>0\) so that \(|r_j(u)-r_j(t)|<\varepsilon\) whenever \(|u-t|<\delta\). If \(0<|h|<\delta\), every \(u\) between \(t\) and \(t+h\) satisfies this bound. Therefore, including when \(h<0\) and the integral has reversed orientation,
Thus the \(j\)th component’s difference quotient tends to \(r_j(t)\), so \(R_j'(t)=r_j(t)\). This holds for every component \(j\), which proves \(\mathbf R'(t)=\mathbf r(t)\).
The Fundamental Theorem: Evaluating a Definite Integral
The other part of the theorem lets us evaluate a definite vector integral without finding a limit of sums. A vector function \(\mathbf F\) is an antiderivative of \(\mathbf r\) on an interval if \(\mathbf F'(t)=\mathbf r(t)\) throughout that interval. Equivalently, each component \(F_j\) is an antiderivative of the corresponding component \(r_j\).
Here is a justification from the variable-upper-endpoint result. Define \(\mathbf H(t)=\int_a^t\mathbf r(u)\,du\) on \(U\). Since \(\mathbf r\) is continuous, the result just proved gives \(\mathbf H'(t)=\mathbf r(t)\). Hence \((\mathbf F-\mathbf H)'(t)=\mathbf 0\). Each component of \(\mathbf F-\mathbf H\) has zero derivative, so by the Mean Value Theorem each is constant on the interval between \(a\) and \(b\). Since \(\mathbf H(a)=\mathbf 0\), evaluating \(\mathbf F-\mathbf H\) at the two endpoints gives \(\mathbf H(b)=\mathbf F(b)-\mathbf F(a)\). The definition of \(\mathbf H\) now yields the stated formula.
The endpoint subtraction is essential: a definite integral records the net change in an antiderivative from the lower endpoint to the upper endpoint. Both endpoints must belong to an interval where the antiderivative relation holds.
Worked Examples
Worked Example: Integrating a Vector with Trigonometric and Exponential Components
Evaluate \(\int_0^{\pi/2}\langle 3t^2-2,4\cos t,e^t\rangle\,dt\).
First, find an antiderivative component by component:
Check it by differentiating each component:
Now evaluate the antiderivative at the upper endpoint:
At the lower endpoint, \(\mathbf F(0)=\langle 0,0,1\rangle\). Subtract componentwise:
Check: differentiating the antiderivative gives the original integrand, and the endpoint subtraction gives the componentwise values \(\int_0^{\pi/2}(3t^2-2)\,dt=\frac{\pi^3}{8}-\pi\), \(\int_0^{\pi/2}4\cos t\,dt=4\), and \(\int_0^{\pi/2}e^t\,dt=e^{\pi/2}-1\).
Worked Example: Evaluating a Polynomial Vector Integral
Evaluate \(\int_{-1}^{2}\langle 2t+1,3t^2,-4\rangle\,dt\).
Integrate each component to obtain an antiderivative:
Differentiate to check the result:
Evaluate at the upper endpoint:
Evaluate at the lower endpoint:
Subtract the lower-endpoint vector from the upper-endpoint vector:
Check componentwise: \(\int_{-1}^{2}(2t+1)\,dt=6\), \(\int_{-1}^{2}3t^2\,dt=9\), and \(\int_{-1}^{2}-4\,dt=-12\). These agree with the components found by the theorem.
Worked Example: Differentiating an Accumulated Vector Integral
For \(t\in(0,3)\), define \(\mathbf A(t)=\int_1^t\langle u^2,\cos u\rangle\,du\). Find \(\mathbf A'(2)\), and calculate \(\mathbf A(2)\).
The integrand \(\mathbf r(u)=\langle u^2,\cos u\rangle\) is continuous on the open interval \((0,3)\), and \(1\) belongs to that interval. The variable-upper-endpoint theorem therefore applies at every \(t\in(0,3)\):
To calculate \(\mathbf A(2)\), integrate each component from \(1\) to \(2\):
Check the derivative by differentiating the evaluated accumulation function \(\mathbf A(t)=\langle (t^3-1)/3,\sin t-\sin 1\rangle\). This gives \(\mathbf A'(t)=\langle t^2,\cos t\rangle\), and substitution of \(t=2\) gives \(\langle 4,\cos 2\rangle\), as required.
A Common Endpoint Mistake
A frequent error is to evaluate the antiderivative at the upper endpoint and stop, forgetting to subtract its value at the lower endpoint. For example, consider \(\int_1^3\langle 2t,3\rangle\,dt\). An incorrect calculation finds \(\mathbf F(t)=\langle t^2,3t\rangle\) and reports \(\mathbf F(3)=\langle 9,9\rangle\) as the integral. That is only the upper-endpoint value, not the net change.
The correction is to evaluate both endpoints and subtract:
Checking directly gives \(\int_1^3 2t\,dt=9-1=8\) and \(\int_1^3 3\,dt=9-3=6\). The corrected vector is \(\langle 8,6\rangle\).
What to Keep in Mind
The Fundamental Theorem for vector integrals does not require a new integration method for vectors. It packages the scalar theorem component by component. For a variable upper endpoint, continuity on an open interval guarantees the derivative equality throughout that interval. For a definite integral, a vector antiderivative gives the upper-endpoint value minus the lower-endpoint value.
Check Your Understanding
Use componentwise integration and the endpoint conditions carefully.
- If \(\mathbf r(t)=\langle 2t,1\rangle\), find \(\int_0^2\mathbf r(t)\,dt\) using a vector antiderivative.
- For continuous \(\mathbf r\) on an open interval \(I\), define \(\mathbf R(t)=\int_a^t\mathbf r(u)\,du\), where \(a\in I\). What is \(\mathbf R'(t)\), and on what inputs is the theorem asserted?
- Find an antiderivative of \(\langle \cos t,2t\rangle\), then verify it by differentiation.
- What endpoint subtraction is missing from the incorrect claim that \(\int_1^4\langle 1,2t\rangle\,dt=\langle 4,16\rangle\)?
- In one sentence, explain why the definite integral of a vector-valued function is also a vector.
