Differentiating an Accumulation Vector Integral

Accumulation and a Moving Upper Limit

A definite integral with a fixed upper endpoint gives a single accumulated vector. If the upper endpoint moves as a parameter changes, the integral itself becomes a vector-valued function. Differentiating that function tells us how quickly the accumulated vector changes as the endpoint moves.

The rule is the vector counterpart of the fundamental theorem of calculus: differentiate each component of the integral, then account for the rate at which its upper endpoint changes. The endpoint and continuity assumptions matter; without them, the integral may not even be defined.

Definition and Conditions

Let $J$ be an interval, let $\mathbf{f}:J\to\mathbb{R}^n$ be continuous, and choose a fixed number $a\in J$. The accumulation vector function with upper endpoint $x\in J$ is defined by integrating each component of $\mathbf{f}$:

\mathbf{A}(x) = \int_a^x \mathbf{f}(u)\,du = \left\langle \int_a^x f_1(u)\,du,\ldots, \int_a^x f_n(u)\,du \right\rangle, \qquad \mathbf{f}(u)=\langle f_1(u),\ldots,f_n(u)\rangle.

Because $J$ is an interval, the segment between $a$ and $x$ lies in $J$. Continuity of $\mathbf{f}$ therefore ensures that each component integral is defined for these endpoints.

Theorem: If $\mathbf{f}:J\to\mathbb{R}^n$ is continuous on an interval $J$, and $a\in J$, then the accumulation function $\mathbf{A}(x)=\int_a^x\mathbf{f}(u)\,du$ is differentiable for $x$ in the interior of $J$, and $$ \mathbf{A}'(x)=\mathbf{f}(x). $$

There is also a useful form when the upper endpoint depends on another variable. Let $K$ be an interval, and let $g:K\to J$ be differentiable. Define

\mathbf{G}(t)=\int_a^{g(t)}\mathbf{f}(u)\,du.

For this definition and the differentiation rule to apply, $\mathbf{f}$ must be continuous on an interval containing both $a$ and every value $g(t)$ under consideration. In particular, $a$ and those values must belong to the domain on which $\mathbf{f}$ is defined.

Theorem: Under these conditions, at each $t\in K$ for which $g(t)$ is an interior point of $J$, $$ \mathbf{G}'(t)=\mathbf{f}(g(t))\,g'(t). $$

Here $g'(t)$ is a scalar, so the right-hand side means scalar multiplication of the vector $\mathbf{f}(g(t))$. The two parts of the rule have distinct roles: evaluate the integrand at the current upper endpoint, then multiply by the rate at which that endpoint changes.

Why the Rule Works

The justification follows from the scalar fundamental theorem of calculus and the chain rule. Write the components of $\mathbf{f}$ as $f_1,\ldots,f_n$. By the definition of the vector integral, the $i$-th component of $\mathbf{G}(t)$ is the scalar function

G_i(t)=\int_a^{g(t)} f_i(u)\,du.

Since $f_i$ is continuous on $J$, the scalar fundamental theorem says that the function $H_i(x)=\int_a^x f_i(u)\,du$ has derivative $H_i'(x)=f_i(x)$ at interior points of $J$. Also, $G_i(t)=H_i(g(t))$. Applying the scalar chain rule gives the derivative of each component:

G_i'(t) = H_i'(g(t))g'(t) = f_i(g(t))g'(t), \qquad i=1,\ldots,n.

A vector-valued function is differentiable componentwise, and its derivative consists of the derivatives of its components. Collecting the component equations therefore gives

\mathbf{G}'(t) = \langle G_1'(t),\ldots,G_n'(t)\rangle = \langle f_1(g(t))g'(t),\ldots,f_n(g(t))g'(t)\rangle = \mathbf{f}(g(t))g'(t).

This argument explains why the formula requires continuity of the integrand and differentiability of the upper endpoint: continuity gives the fundamental theorem for each component, and differentiability of $g$ permits the chain rule.

Worked Examples

Worked Example: A Polynomial Upper Endpoint

Let $\mathbf{f}(u)=\langle u^2,\sin u,e^u\rangle$, and define $\mathbf{A}(t)=\int_0^{t^2+1}\mathbf{f}(u)\,du$ for real $t$. Find $\mathbf{A}'(t)$, then evaluate it at $t=1$.

The integrand is continuous for every real $u$. The fixed lower endpoint $0$ and every upper endpoint $t^2+1$ lie in $J=\mathbb{R}$. Set $g(t)=t^2+1$, so $g'(t)=2t$. The rule gives

\mathbf{A}'(t) = \mathbf{f}(t^2+1)(2t) = \left\langle (t^2+1)^2,\sin(t^2+1),e^{t^2+1} \right\rangle(2t).

Multiplying each component by $2t$,

\mathbf{A}'(t) = \left\langle 2t(t^2+1)^2,\, 2t\sin(t^2+1),\, 2t e^{t^2+1} \right\rangle.

At $t=1$, the upper endpoint is $1^2+1=2$, and $g'(1)=2$. Thus

\mathbf{A}'(1) = \langle 2^2,\sin 2,e^2\rangle(2) = \langle 8,2\sin 2,2e^2\rangle.

Checking componentwise, the integrand at the endpoint $u=2$ is $\langle 4,\sin 2,e^2\rangle$; multiplying by the endpoint rate $2$ gives exactly $\langle 8,2\sin 2,2e^2\rangle$.

Worked Example: Checking the Domain of the Integrand

Let $\mathbf{f}(u)=\langle\sqrt{u},1/u,u^3\rangle$, and define $\mathbf{G}(t)=\int_1^{3t-1}\mathbf{f}(u)\,du$ for $t>1/3$. Find $\mathbf{G}'(1)$.

The components of $\mathbf{f}$ are continuous on $J=(0,\infty)$. The lower endpoint $1$ belongs to $J$, and $3t-1>0$ whenever $t>1/3$, so every upper endpoint under consideration is also in $J$. Here $g(t)=3t-1$, so $g'(t)=3$. At $t=1$, $g(1)=2$, and

\mathbf{G}'(1) = \mathbf{f}(2)\,3 = \left\langle \sqrt{2},\frac12,8 \right\rangle 3 = \left\langle 3\sqrt{2},\frac32,24 \right\rangle.

For a componentwise check, the endpoint values of the three integrand components are $\sqrt{2}$, $1/2$, and $2^3=8$. The upper endpoint increases at rate $3$, so the corresponding rates of accumulation are $3\sqrt{2}$, $3/2$, and $24$, as found.

Worked Example: A Trigonometric Upper Endpoint

Let $\mathbf{f}(u)=\langle u,u^2\rangle$, and define $\mathbf{G}(t)=\int_{-1}^{\sin t}\mathbf{f}(u)\,du$. Find $\mathbf{G}'(t)$ and then evaluate it at $t=\pi/3$.

The integrand is continuous on $J=\mathbb{R}$. The lower endpoint $-1$ and every value of $\sin t$ lie in this interval. Since $g(t)=\sin t$, its derivative is $g'(t)=\cos t$. Therefore,

\mathbf{G}'(t) = \mathbf{f}(\sin t)\cos t = \langle\sin t,(\sin t)^2\rangle\cos t = \langle \sin t\cos t,\sin^2 t\cos t\rangle.

At $t=\pi/3$, $\sin(\pi/3)=\sqrt{3}/2$ and $\cos(\pi/3)=1/2$. Substitution gives

\mathbf{G}'\left(\frac{\pi}{3}\right) = \left\langle \frac{\sqrt{3}}{2}\cdot\frac12,\, \left(\frac{\sqrt{3}}{2}\right)^2\cdot\frac12 \right\rangle = \left\langle \frac{\sqrt{3}}{4},\frac{3}{8} \right\rangle.

Indeed, the integrand at the current endpoint is $\mathbf{f}(\sqrt{3}/2)=\langle\sqrt{3}/2,3/4\rangle$, and the endpoint rate is $1/2$. Their scalar product is $\langle\sqrt{3}/4,3/8\rangle$, confirming both components.

When the Upper Endpoint Stops or Reverses

The formula also clarifies what happens when the endpoint is momentarily stationary. If $g'(t_0)=0$, then $\mathbf{G}'(t_0)=\mathbf{0}$, provided the hypotheses hold. At that parameter value the endpoint has zero instantaneous rate of change, so the accumulated vector has zero instantaneous rate of change as well.

If $g'(t_0)<0$, the derivative formula still applies. The negative scalar records that the upper endpoint is moving in the decreasing direction. No separate sign adjustment is needed: the integral’s orientation is already accounted for by its endpoints, and the chain factor $g'(t_0)$ supplies the direction of change.

It is important not to confuse $\mathbf{G}'(t)$ with $\mathbf{f}(g(t))$ when the endpoint is a function of $t$. The latter gives the integrand at the endpoint; the former also depends on how fast that endpoint moves.

Common Mistakes and Corrections

A frequent error is to substitute the moving endpoint into the integrand but omit the derivative of the endpoint. For example, take $\mathbf{f}(u)=\langle u,1\rangle$ and $\mathbf{G}(t)=\int_0^{t^2}\mathbf{f}(u)\,du$. An incorrect calculation would be

\text{Incorrect:}\qquad \mathbf{G}'(t)=\mathbf{f}(t^2)=\langle t^2,1\rangle.

This omits the endpoint rate. Since $g(t)=t^2$ and $g'(t)=2t$, the correct calculation is

\mathbf{G}'(t) = \mathbf{f}(t^2)(2t) = \langle t^2,1\rangle(2t) = \langle 2t^3,2t\rangle.

At $t=2$, the incorrect result is $\langle4,1\rangle$, while the correct result is $\langle16,4\rangle$. The endpoint is moving at rate $4$ at $t=2$, so each component’s rate of accumulation must include that factor.

A separate error is to check continuity only near the values of $g(t)$ and forget the fixed lower endpoint $a$. For instance, an integrand defined only on $(0,1)$ cannot be integrated from $a=2$, even if $g(t)=1/2$. The integral would require values of the integrand between $1/2$ and $2$, outside its domain. To use the theorem, the integrand must be continuous on an interval containing both the lower endpoint and the upper endpoints being considered.

Why Accumulation Derivatives Matter

An accumulation function represents a running total whose endpoint changes with a parameter. The differentiation rule converts that running total into local information: the instantaneous change equals the vector being accumulated at the current endpoint, scaled by the endpoint’s rate of change. Componentwise integration makes this interpretation precise for each coordinate while preserving the vector structure.

Before applying the rule, check three things: the integrand is continuous on an interval containing the fixed lower endpoint and the relevant upper endpoints; the upper endpoint is differentiable; and the final answer includes its derivative as a scalar factor. These checks establish both that the integral is defined and that the differentiation argument applies.

Check Your Understanding

Use the componentwise fundamental theorem and the chain rule to answer these questions.

State the continuity and endpoint conditions needed for differentiating an accumulation vector integral.
If $\mathbf{A}(x)=\int_a^x\mathbf{f}(u)\,du$ and $\mathbf{f}$ is continuous, what is $\mathbf{A}'(x)$?
For $\mathbf{G}(t)=\int_a^{g(t)}\mathbf{f}(u)\,du$, explain the role of $g'(t)$ in the derivative.
For $\mathbf{G}(t)=\int_0^{t^2}\langle u,1\rangle\,du$, find $\mathbf{G}'(1)$.
Why is continuity near the upper endpoints alone insufficient if the fixed lower endpoint is outside the domain of the integrand?