Evaluating Definite Vector Integrals with Antiderivatives

From an Antiderivative to a Definite Vector Integral

A definite vector integral accumulates the change in each component of a vector-valued function over an interval. When an antiderivative is available, the fundamental theorem of calculus turns that accumulation into an endpoint calculation: evaluate the antiderivative at the upper bound, evaluate it at the lower bound, and subtract.

The procedure resembles evaluating a definite scalar integral, but the result is a vector. Each component must be integrated and evaluated in its own coordinate. Keeping those components organized is the main practical skill in this tutorial.

Definition: If $\mathbf{f}:[a,b]\to\mathbb{R}^n$ has component functions $f_1,\ldots,f_n$, its definite integral is defined componentwise: $$ \int_a^b \mathbf{f}(t)\,dt = \left\langle \int_a^b f_1(t)\,dt,\ldots,\int_a^b f_n(t)\,dt \right\rangle. $$

In particular, the integral has the same number of components as $\mathbf{f}$. The components are scalar definite integrals; they are assembled into one vector after they are evaluated.

The Componentwise Fundamental Theorem

A vector antiderivative of $\mathbf{f}$ on an interval is a vector-valued function $\mathbf{F}$ whose derivative is $\mathbf{f}$ there. In other words, if $\mathbf{F}(t)=\langle F_1(t),\ldots,F_n(t)\rangle$, then $F_i'(t)=f_i(t)$ for every component $i$. As with vector derivatives, this condition is checked component by component.

Theorem: Suppose each component $f_i$ is continuous on $[a,b]$, and $\mathbf{F}$ is an antiderivative of $\mathbf{f}$ on that interval. Then $$ \int_a^b \mathbf{f}(t)\,dt = \mathbf{F}(b)-\mathbf{F}(a). $$ The subtraction on the right is vector subtraction: subtract corresponding components.

Continuity of each component ensures its definite integral exists. The antiderivative condition allows the scalar fundamental theorem of calculus to be applied to each component. The formula does not require $\mathbf{F}(a)$ to be the zero vector. Its value must be subtracted just as carefully as the value at the upper bound.

Why Endpoint Subtraction Works

Write $\mathbf{F}(t)=\langle F_1(t),\ldots,F_n(t)\rangle$ and $\mathbf{f}(t)=\langle f_1(t),\ldots,f_n(t)\rangle$. By the definition of a vector antiderivative, $F_i'(t)=f_i(t)$ for each $i$. The scalar fundamental theorem says that the integral of each $f_i$ is the change in $F_i$ between the endpoints.

\int_a^b f_i(t)\,dt=F_i(b)-F_i(a) \qquad\text{for }i=1,\ldots,n.

Putting these component equalities together gives the vector identity:

\begin{aligned} \int_a^b \mathbf{f}(t)\,dt &= \left\langle \int_a^b f_1(t)\,dt,\ldots,\int_a^b f_n(t)\,dt \right\rangle\\ &= \left\langle F_1(b)-F_1(a),\ldots,F_n(b)-F_n(a) \right\rangle\\ &= \mathbf{F}(b)-\mathbf{F}(a). \end{aligned}

This proves the vector formula using only the scalar fundamental theorem and the componentwise definitions of vector integration and differentiation. The endpoint rule works because it holds in every coordinate, not because vectors require a separate scalar-like notion of antiderivative.

A Reliable Evaluation Procedure

For a definite vector integral, keep the component calculations together but distinct. First find one antiderivative for each component and assemble them into $\mathbf{F}$. Then evaluate the entire vector at $b$ and at $a$. Finally subtract the lower-endpoint vector from the upper-endpoint vector.

Find the antiderivative.
Integrate each component of $\mathbf{f}(t)$ with respect to $t$, and write the results as $\mathbf{F}(t)$.

Check the derivative.
Differentiate each component of $\mathbf{F}(t)$ to confirm that $\mathbf{F}'(t)=\mathbf{f}(t)$.

Substitute the bounds.
Calculate $\mathbf{F}(b)$ and $\mathbf{F}(a)$, keeping all components in their original order.

Subtract endpoint values.
Compute $\mathbf{F}(b)-\mathbf{F}(a)$ component by component.

An arbitrary constant vector could be added to $\mathbf{F}$, just as an arbitrary constant could be added to a scalar antiderivative. It does not change the endpoint difference: if $\mathbf{C}$ is constant, then $(\mathbf{F}(b)+\mathbf{C})-(\mathbf{F}(a)+\mathbf{C})=\mathbf{F}(b)-\mathbf{F}(a)$. Therefore, it is usually simplest to use zero constants in the component antiderivatives.

Worked Examples

Worked Example: Polynomial Components

Evaluate $\int_0^2 \langle 3t^2-2,\,4t,\,1\rangle\,dt$.

Integrate each component with respect to $t$:

\mathbf{F}(t) = \left\langle \int (3t^2-2)\,dt,\, \int 4t\,dt,\, \int 1\,dt \right\rangle = \langle t^3-2t,\,2t^2,\,t\rangle.

Differentiate to check the antiderivative:

\mathbf{F}'(t) = \langle 3t^2-2,\,4t,\,1\rangle.

Evaluate at the upper bound $t=2$:

\mathbf{F}(2) = \langle 2^3-2(2),\,2(2^2),\,2\rangle = \langle 8-4,\,8,\,2\rangle = \langle 4,\,8,\,2\rangle.

Evaluate at the lower bound $t=0$:

\mathbf{F}(0) = \langle 0^3-2(0),\,2(0^2),\,0\rangle = \langle 0,\,0,\,0\rangle.

Subtract the lower value from the upper value:

\int_0^2 \langle 3t^2-2,\,4t,\,1\rangle\,dt = \langle 4,\,8,\,2\rangle-\langle 0,\,0,\,0\rangle = \langle 4,\,8,\,2\rangle.

The derivative check confirms that each component of the chosen antiderivative gives the corresponding integrand component, so the endpoint calculation applies.

Worked Example: Trigonometric and Exponential Components

Evaluate $\int_0^{\pi/2}\langle \cos t,\,-\sin t,\,2e^{2t}\rangle\,dt$.

Integrating componentwise gives the antiderivative $\mathbf{F}(t)=\langle \sin t,\,\cos t,\,e^{2t}\rangle$. In particular, the derivative of $\cos t$ is $-\sin t$, and the chain rule gives $\frac{d}{dt}e^{2t}=2e^{2t}$. Thus

\mathbf{F}'(t)=\langle \cos t,\,-\sin t,\,2e^{2t}\rangle.

At the upper endpoint, use $\sin(\pi/2)=1$, $\cos(\pi/2)=0$, and $e^{2(\pi/2)}=e^\pi$:

\mathbf{F}\left(\frac{\pi}{2}\right) = \left\langle 1,\,0,\,e^\pi\right\rangle.

At the lower endpoint, $\sin 0=0$, $\cos 0=1$, and $e^0=1$:

\mathbf{F}(0)=\langle 0,\,1,\,1\rangle.

Now subtract corresponding components:

\begin{aligned} \int_0^{\pi/2}\langle \cos t,\,-\sin t,\,2e^{2t}\rangle\,dt &= \left\langle 1,\,0,\,e^\pi\right\rangle - \langle 0,\,1,\,1\rangle\\ &= \langle 1-0,\,0-1,\,e^\pi-1\rangle\\ &= \langle 1,\,-1,\,e^\pi-1\rangle. \end{aligned}

As a check, differentiating the three entries of $\mathbf{F}$ reproduces $\cos t$, $-\sin t$, and $2e^{2t}$, respectively. The negative second component of the answer is consistent with the endpoint change in $\cos t$, which decreases from $1$ to $0$.

Worked Example: Logarithmic Components

Evaluate $\int_0^1\left\langle \frac{1}{t+1},\,\frac{2t}{t^2+1},\,3\right\rangle\,dt$.

On $[0,1]$, $t+1$ is positive and $t^2+1$ is positive, so the logarithmic antiderivatives are defined throughout the interval. A componentwise antiderivative is

\mathbf{F}(t) = \langle \ln(t+1),\,\ln(t^2+1),\,3t\rangle.

Check each derivative. The chain rule yields $\frac{d}{dt}\ln(t+1)=\frac{1}{t+1}$ and $\frac{d}{dt}\ln(t^2+1)=\frac{2t}{t^2+1}$; also, $\frac{d}{dt}(3t)=3$. Therefore, $\mathbf{F}'(t)$ is the stated integrand.

At $t=1$, the three components are $\ln 2$, $\ln 2$, and $3$:

\mathbf{F}(1)=\langle \ln 2,\,\ln 2,\,3\rangle.

At $t=0$, they are $\ln 1=0$, $\ln 1=0$, and $0$:

\mathbf{F}(0)=\langle 0,\,0,\,0\rangle.

Hence

\begin{aligned} \int_0^1\left\langle \frac{1}{t+1},\,\frac{2t}{t^2+1},\,3\right\rangle\,dt &= \langle \ln 2,\,\ln 2,\,3\rangle-\langle 0,\,0,\,0\rangle\\ &= \langle \ln 2,\,\ln 2,\,3\rangle. \end{aligned}

Differentiating $\mathbf{F}$ returns all three integrand components, verifying the antiderivative used in the evaluation.

A Common Endpoint Mistake

A frequent error is to evaluate $\mathbf{F}$ at the upper bound and report that vector as the integral, without subtracting $\mathbf{F}(a)$. This shortcut works only when $\mathbf{F}(a)=\mathbf{0}$, and that condition should never be assumed without checking.

Worked Example: Correcting a Missing Lower Endpoint

Evaluate $\int_1^2\langle 2t,\,3\rangle\,dt$. A componentwise antiderivative is $\mathbf{F}(t)=\langle t^2,\,3t\rangle$, since $\mathbf{F}'(t)=\langle 2t,\,3\rangle$.

A wrong calculation would stop after evaluating the upper endpoint:

\mathbf{F}(2)=\langle 2^2,\,3(2)\rangle=\langle 4,\,6\rangle.

But this omits the lower-endpoint value. The correct calculation is

\mathbf{F}(1)=\langle 1^2,\,3(1)\rangle=\langle 1,\,3\rangle, \qquad \mathbf{F}(2)-\mathbf{F}(1) = \langle 4-1,\,6-3\rangle = \langle 3,\,3\rangle.

The derivative check is $\frac{d}{dt}\langle t^2,3t\rangle=\langle 2t,3\rangle$, and the corrected result uses both endpoint values as required.

What the Result Means

The result of a definite vector integral is a vector of componentwise accumulated changes. The endpoint formula makes this explicit: the $i$th component of the answer is $F_i(b)-F_i(a)$, the change in the $i$th component of the antiderivative over the interval. A component may be positive, negative, or zero, depending on its net change.

This coordinate-by-coordinate meaning is also why it is important not to replace the vector integral with a single scalar quantity such as the magnitude of the integrand. The fundamental theorem evaluates the vector integral by applying the scalar theorem to each component. The result is the vector of those signed component integrals.

Before accepting an answer, make two checks: confirm that differentiating the proposed $\mathbf{F}$ reproduces every component of $\mathbf{f}$, and confirm that the final subtraction is upper endpoint minus lower endpoint in every coordinate. These checks catch most errors in evaluating definite vector integrals with antiderivatives.

Check Your Understanding

Use componentwise antiderivatives and the endpoint formula to answer these questions.

Find an antiderivative of $\mathbf{f}(t)=\langle 6t,\,\cos t\rangle$, then evaluate $\int_0^1\mathbf{f}(t)\,dt$.
For $\mathbf{F}(t)=\langle t^2,\,e^t\rangle$, what is $\mathbf{F}(b)-\mathbf{F}(a)$ in component notation?
Why must each component of $\mathbf{F}$ be differentiated when checking that it is an antiderivative of $\mathbf{f}$?
Evaluate $\int_1^3\langle 1,\,2t\rangle\,dt$, showing both endpoint values before subtracting.
State the continuity condition that ensures the component integrals in the theorem exist on $[a,b]$.