Calculus I · Unit 3A · lesson

Accumulation Functions

Learn Accumulation Functions through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.

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Section overview

The Fundamental Theorem of Calculus

What this section is building

Learn Accumulation Functions through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.

Notice

Differentiating a running total recovers its current integrand, while evaluating an accumulated total subtracts antiderivative endpoint values.

Decide

Separate FTC Part I, FTC Part II, net change, and variable-bound chain-rule tasks before manipulating notation.

Avoid

Forgetting a chain-rule factor at a variable bound, reversing endpoint subtraction, or adding +C to a definite value.

Concept

Learning objectives

Interpret a variable-upper-limit integral as a new function and reason about its increase, decrease, and units.

Accumulation Functions

Explanation

An integral can define a new function

When the upper limit varies, $A(x)=\int_a^x f(t)\,dt$ records how much signed accumulation has occurred by the time the input reaches $x$ . The dummy variable $t$ is used inside the integral so that $x$ can remain free to control the endpoint. As $x$ moves, the accumulated region grows, shrinks, or gains negative contribution depending on the sign of $f$ .

The graph of $A$ can often be predicted directly from the graph of $f$ . Where $f$ is positive, $A$ increases; where $f$ is negative, $A$ decreases; where $f=0$ , $A$ has a horizontal tangent. This makes accumulation functions a bridge between geometric area, net change, and derivative behavior, setting up the Fundamental Theorem rather than appearing as an isolated notation trick.

Fix a starting point $a$ and define

A(x)=\int_a^x f(t)\,dt.

The variable $t$ is a dummy variable inside the integral; $x$ determines where accumulation stops.

If $f>0$ , moving $x$ to the right adds positive contributions and $A$ increases. If $f<0$ , $A$ decreases. The value $A(x)$ is a total; the value $f(x)$ is the local rate at which that total changes.

Worked example

Build an accumulation function directly

For $f(t)=2t$ and $a=1$ ,

A(x)=\int_1^x2t\,dt=x^2-1.

Thus $A(1)=0$ , $A(2)=3$ , and $A'(x)=2x=f(x)$ .

Interactive checku3a-accumulation-01

For $A(x)=\int_3^x f(t)\,dt$ , what is $A(3)$ ?

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Show hint

The interval has zero width.

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**Visual study**
Moving upper bound and accumulated area. Show A(x)=int a x f(t)dt changing as x moves.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Accumulation Functions

Explanation

An integral can define a new function

Fix a starting point

a

and define

A(x)=\int_a^x f(t)\,dt.

The variable

t

is a dummy variable inside the integral;

x

determines where accumulation stops.

f>0

, moving

x

to the right adds positive contributions and

A

increases. If

f<0

A

decreases. The value

A(x)

is a total; the value

f(x)

is the local rate at which that total changes.

Worked example

Build an accumulation function directly

For $f(t)=2t$ and $a=1$ ,

A(x)=\int_1^x2t\,dt=x^2-1.

Thus $A(1)=0$ , $A(2)=3$ , and $A'(x)=2x=f(x)$ .

Interactive checku3a-accumulation-01

For $A(x)=\int_3^x f(t)\,dt$ , what is $A(3)$ ?

Your work stays on this device. No account or AI grader is used.

Show hint

The interval has zero width.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Read this graph as text

Moving upper bound and accumulated area. As x moves, the shaded signed area updates and a point traces A(x). Show A(x)=int a x f(t)dt changing as x moves. Preserve sign when the integrand is below the axis.

Every relationship in moving upper bound and accumulated area uses written labels together with distinct line styles, markers, or fill patterns; color is never the only carrier of meaning.

Why it matters: Show A(x)=int a x f(t)dt changing as x moves.

Visual study

Moving upper bound and accumulated area. Show A(x)=int a x f(t)dt changing as x moves.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Accumulation Functions

Related resources

Original instruction with traceable references.

Accumulation Functions

Related resources

Original instruction with traceable references.