🎯⭐ INTERACTIVE LESSON

Accumulation Functions

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Accumulation Functions - Complete Interactive Lesson

Part 1: The Accumulation Concept

Accumulation Functions

Part 1 of 7 — The Accumulation Concept

The Accumulation Concept
Reading Graphs of $f$ to Analyze $F$
FTC Part 1 with Chain Rule
Net Change Applications
Average Value
Practice Workshop
Comprehensive Assessment

What is an Accumulation Function?

$\boxed{F(x) = \int_a^x f(t)\,dt}$

$F(x)$ measures how much $f$ has accumulated from $a$ to $x$ .

Key Properties at a Glance

Property	Formula	Interpretation
Value at start	$F(a) = 0$	Nothing accumulated yet
Derivative	$F'(x) = f(x)$	Rate of accumulation = integrand
$F$ increasing	$f(x) > 0$	Positive rate → accumulating
$F$ decreasing	$f(x) < 0$	Negative rate → depleting
$F$ has local max	$f$ changes $+ \to -$	Rate switches from growth to decline
$F$ has local min	$f$ changes $- \to +$	Rate switches from decline to growth
$F$ concave up	$f'(x) > 0$ ( $f$ increasing)	Rate is accelerating
$F$ concave down	$f'(x) < 0$ ( $f$ decreasing)	Rate is decelerating

Key Fact: The accumulation function connects the three layers: $F$ , $F' = f$ , and $F'' = f'$ .

Worked Example

Let $F(x) = \int_0^x (2t - 4)\,dt$ .

Quantity	Computation	Result
$F(0)$	$\int_0^0 = 0$	$0$
$F(2)$	$[t^2-4t]_0^2 = 4-8$	$-4$
$F(3)$	$[t^2-4t]_0^3 = 9-12$	$-3$
$F(4)$	$[t^2-4t]_0^4 = 16-16$	$0$
$F'(3)$	$f(3) = 2(3)-4$	$2$

Interpretation: At $x = 2$ , the function $f(t) = 2t-4$ crosses zero (changes from negative to positive). So $F$ has a local minimum at $x = 2$ .

$\text{Local min at } x = 2: \quad f(2) = 0, \quad f \text{ changes } - \to +$

AP Tip: On the exam, you must justify extrema by showing $f$ changes sign, not just that $f = 0$ .

Accumulation Functions 🎯

Let $g(x) = \int_1^x f(t)\,dt$ where $f$ is continuous.

Connect $f$ and $F$ . 🔍

Compute an accumulation value. ✍️

Key Takeaways — Part 1

Concept	Key Point
$F(x) = \int_a^x f(t)\,dt$	Accumulates from $a$ to $x$
$F(a) = 0$	Always starts at zero
$F' = f$	Rate of accumulation = integrand
$F'' = f'$	Concavity of $F$ = slope of $f$
Extrema of $F$	Where $f$ changes sign

Up Next: Part 2 — Reading Graphs of $f$ to Analyze $F$ .

Part 2: Reading Graphs of f to Analyze F

Accumulation Functions

Part 2 of 7 — Reading Graphs of $f$ to Analyze $F$

The Most Tested AP Skill

Given the graph of $f$ , determine everything about $F(x) = \int_a^x f(t)\,dt$ .

$\boxed{f \text{ (graph given)} \xrightarrow{\text{integrate}} F \text{ (properties to find)}}$

Complete Translation Guide

Given about $f$	Conclude about $F$
$f(c) = 0$	$F'(c) = 0$ (critical point)
$f > 0$ on interval	$F$ increasing
$f < 0$ on interval	$F$ decreasing
$f$ changes $+ \to -$	$F$ has local max
$f$ changes $- \to +$	$F$ has local min
$f$ increasing	$F$ concave up ( $F'' = f' > 0$ )
$f$ decreasing	$F$ concave down ( $F'' = f' < 0$ )
$f$ has local max or min	$F$ has inflection point
Area under $f$ above axis	Positive contribution to $F$
Area under $f$ below axis	Negative contribution to $F$

AP Tip: Always write $g' = f$ and $g'' = f'$ at the top of your work. This prevents confusion between the layers.

Computing $F(x)$ from Geometric Areas

When $f$ is piecewise linear, compute $F(x)$ using geometric shapes:

Shape	Area Formula
Rectangle	$\text{base} \times \text{height}$
Triangle	$\frac{1}{2} \times \text{base} \times \text{height}$
Trapezoid	$\frac{1}{2}(b_1 + b_2) \times h$
Semicircle	$\frac{1}{2}\pi r^2$

Worked Example

$f$ is piecewise linear: $f(0)=2$ , $f(2)=2$ , $f(4)=-2$ , $f(6)=0$ .

Interval	Shape	Signed Area	Running Total $F(x)$
$[0,2]$	Rectangle	$2 \times 2 = +4$	$F(2) = 4$
$[2,3]$	Triangle	$\frac{1}{2}(1)(2) = +1$	$F(3) = 5$
$[3,4]$	Triangle	$\frac{1}{2}(1)(-2) = -1$	Lost: $f=0$ at $x=3$
$[2,4]$	Trapezoid	$\frac{1}{2}(2)(2+(-2)) = 0$	$F(4) = 4$
$[4,6]$	Triangle	$\frac{1}{2}(2)(-2) = -2$	$F(6) = 2$

Note: $f$ crosses zero at $x = 3$ (linear from $2$ to $-2$ ). $F$ has its maximum at $x = 3$ .

Graph Analysis 🎯

Suppose $f$ is piecewise linear: $f(0)=2$ , $f(2)=2$ , $f(4)=-2$ , $f(6)=0$ . Let $g(x) = \int_0^x f(t)\,dt$ .

Match behavior to conclusions. 🔍

Compute from areas. ✍️

Key Takeaways — Part 2

Concept	Key Point
$f > 0$	$F$ increasing, positive area
$f < 0$	$F$ decreasing, negative area
$f$ crosses zero	$F$ has local extremum
$f$ constant	$F$ is linear
$f$ linear	$F$ is quadratic

Up Next: Part 3 — FTC Part 1 with Chain Rule.

Part 3: FTC Part 1 with Chain Rule Review

Accumulation Functions

Part 3 of 7 — FTC Part 1 with Chain Rule

Standard FTC Part 1

$\boxed{\frac{d}{dx}\int_a^x f(t)\,dt = f(x)}$

Chain Rule Extension

$\boxed{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)}$

All Variations

Form	Result	Key Step
$\frac{d}{dx}\int_a^x f(t)\,dt$	$f(x)$	Direct FTC
$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt$	$f(g(x)) \cdot g'(x)$	Chain rule on upper
$\frac{d}{dx}\int_x^b f(t)\,dt$	$-f(x)$	Flip limits, negate
$\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt$	$f(g(x))g'(x) - f(h(x))h'(x)$	Split and apply to each

Key Fact: The pattern is: evaluate $f$ at the limit, then multiply by the limit's derivative. Subtract the lower limit's contribution.

Worked Examples

Example 1: $\frac{d}{dx}\int_0^{x^3} \sin(t)\,dt$

$= \sin(x^3) \cdot 3x^2$

Example 2: $\frac{d}{dx}\int_0^{\sin x} e^{t^2}\,dt$

$= e^{(\sin x)^2} \cdot \cos x = e^{\sin^2 x} \cos x$

Example 3 (Both limits): $\frac{d}{dx}\int_{2x}^{x^2} e^t\,dt$

Upper: $e^{x^2} \cdot 2x$ . Lower: $e^{2x} \cdot 2$ .

$= 2xe^{x^2} - 2e^{2x}$

Example 4 (Variable in lower limit): $\frac{d}{dx}\int_x^5 \cos(t^2)\,dt$

$= -\cos(x^2) \cdot 1 = -\cos(x^2)$

(Flip: $-\frac{d}{dx}\int_5^x \cos(t^2)\,dt = -\cos(x^2)$ )

FTC with Chain Rule 🎯

Identify the result. 🔍

Apply FTC with chain rule. ✍️

Key Takeaways — Part 3

Variation	Result
Upper limit $x$	$f(x)$
Upper limit $g(x)$	$f(g(x)) \cdot g'(x)$
Lower limit $x$	$-f(x)$
Both limits variable	Upper contribution $-$ lower contribution

Up Next: Part 4 — Net Change Applications.

Part 4: Net Change Applications

Accumulation Functions

Part 4 of 7 — Net Change & Rate Applications

The Net Change Theorem

$\boxed{\int_a^b f'(t)\,dt = f(b) - f(a)}$

The integral of a rate of change gives the net change in the original quantity.

Rate Function	Integral Gives
Velocity $v(t)$	Net displacement: $s(b)-s(a)$
Speed $	v(t)
Population rate $P'(t)$	Net population change
Flow rate $R(t)$ (gal/min)	Net gallons added
Cost rate $C'(x)$	Net cost change

Key Fact: "Net" means signed — positive and negative parts can cancel. "Total" means unsigned — use absolute value.

Displacement vs. Total Distance

Quantity	Formula	Meaning
Displacement	$\int_a^b v(t)\,dt$	Where you end up relative to start
Total distance	$\int_a^b	v(t)

When $v(t)$ changes sign, split the integral at the zeros.

Example: $v(t) = t - 3$ on $[0, 5]$ .

Zero at $t=3$ :

Interval	$\int$	Value
$[0,3]$	$\int_0^3 (t-3)\,dt$	$-\frac{9}{2}$
$[3,5]$	$\int_3^5 (t-3)\,dt$	$2$
Displacement	Sum	$-\frac{9}{2} + 2 = -\frac{5}{2}$
Total distance	Sum of $	\cdot

Rate In / Rate Out Problems

$\boxed{\text{Amount at time } t = \text{Initial} + \int_0^t [R_{\text{in}}(s) - R_{\text{out}}(s)]\,ds}$

Water Tank Example:

Water enters a tank at $R_{\text{in}}(t) = 5 + 4\sin(t^2)$ gal/min and drains at $R_{\text{out}}(t) = 3 + t$ gal/min. Initially 50 gallons.

Question	Setup
Amount at $t=4$	$50 + \int_0^4 [R_{\text{in}}(t) - R_{\text{out}}(t)]\,dt$
When is water increasing?	When $R_{\text{in}}(t) > R_{\text{out}}(t)$
Maximum amount	Find where $R_{\text{in}}(t) = R_{\text{out}}(t)$ and test
Total water entering	$\int_0^4 R_{\text{in}}(t)\,dt$

AP Tip: Rate in/out problems appear on nearly every AP exam Free Response. Always set up the net rate $R_{\text{in}} - R_{\text{out}}$ first.

Net Change Applications 🎯

Interpret the integral. 🔍

Compute net change. ✍️

Key Takeaways — Part 4

Concept	Formula
Net change	$\int_a^b f'(t)\,dt = f(b)-f(a)$
Displacement	$\int_a^b v(t)\,dt$
Total distance	$\int_a^b
Rate in/out	Initial $+ \int_0^t [R_{in}-R_{out}]\,ds$

Up Next: Part 5 — Average Value of a Function.

Part 5: Average Value of a Function

Accumulation Functions

Part 5 of 7 — Average Value of a Function

Average Value Formula

$\boxed{f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx}$

Intuition: The average value is the height of a rectangle with the same base $[a,b]$ that has the same area as the region under $f$ .

Discrete Average	Continuous Average
$\frac{x_1+x_2+\cdots+x_n}{n}$	$\frac{1}{b-a}\int_a^b f(x)\,dx$
Sum divided by count	Integral divided by interval length

Key Fact: The factor $\frac{1}{b-a}$ normalizes the integral. Without it, wider intervals would always give larger "averages."

Mean Value Theorem for Integrals

$\boxed{\text{If } f \text{ is continuous on } [a,b], \text{ then } \exists\, c \in (a,b) \text{ such that } f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx}$

Translation: A continuous function hits its average value at least once.

Worked Example

Find the average value of $f(x) = x^2$ on $[0, 3]$ .

$f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot 9 = 3$

Find $c$ where $f(c) = 3$ : $c^2 = 3 \Rightarrow c = \sqrt{3} \approx 1.732$ .

Step	Computation
Set up	$\frac{1}{3}\int_0^3 x^2\,dx$
Evaluate integral	$\frac{1}{3}\cdot 9 = 3$
Find $c$	$c^2 = 3 \Rightarrow c = \sqrt{3}$
Verify $c \in (0,3)$	$\sqrt{3} \approx 1.73$ ✓

Common Variations on AP Exams

Problem Type	Setup
Average temperature over $[0, 12]$ hours	$\frac{1}{12}\int_0^{12} T(t)\,dt$
Average velocity over $[a,b]$	$\frac{1}{b-a}\int_a^b v(t)\,dt = \frac{s(b)-s(a)}{b-a}$
Average rate of production	$\frac{1}{b-a}\int_a^b R(t)\,dt$
Average value from table data	Use Riemann sum or trapezoidal rule to approximate $\int$

AP Tip: Average velocity $= \frac{\text{displacement}}{\text{time}}$ . This naturally equals $\frac{1}{b-a}\int_a^b v(t)\,dt$ by Net Change Theorem.

Key Distinction

Average velocity	Average speed
$\frac{1}{b-a}\int_a^b v(t)\,dt$	$\frac{1}{b-a}\int_a^b
Uses signed velocity	Uses absolute value
Can be zero or negative	Always $\ge 0$

Average Value 🎯

Average value concepts. 🔍

Compute the average value. ✍️

Key Takeaways — Part 5

Concept	Formula
Average value	$\frac{1}{b-a}\int_a^b f(x)\,dx$
MVT for Integrals	$\exists\,c: f(c) = f_{avg}$
Average velocity	$\frac{s(b)-s(a)}{b-a}$
Reverse: find $\int$ from avg	$\int = f_{avg} \cdot (b-a)$

Up Next: Part 6 — Problem-Solving Workshop.

Part 6: Practice Workshop

Accumulation Functions

Part 6 of 7 — Problem-Solving Workshop

Graph-to-Accumulation Strategy

Given a graph of $f$ and $g(x) = \int_a^x f(t)\,dt$ :

Step	What to Find	How
1	$g(c)$ for specific $c$	Compute signed area from $a$ to $c$
2	$g'(x)$	Equals $f(x)$ by FTC
3	$g$ increasing/decreasing	Where $f > 0$ / $f < 0$
4	Local max/min of $g$	Where $f$ changes sign
5	$g''(x)$	Equals $f'(x)$ (slope of $f$ )
6	Concavity of $g$	Where $f$ is increasing/decreasing
7	Inflection points of $g$	Where $f$ has local extrema

Key Fact: Every property of $g$ is read from $f$ — you never need to find a formula for $g$ .

Worked Example: Piecewise Linear Graph

Suppose $f$ is piecewise linear with vertices at $(0,0)$ , $(2,4)$ , $(5,4)$ , $(7,0)$ , $(9,-2)$ , and $g(x) = \int_0^x f(t)\,dt$ .

Computing $g$ values using geometric areas:

$x$	Shape from previous to $x$	Area	$g(x)$ (running total)
$0$	—	—	$0$
$2$	Triangle: $\frac{1}{2}(2)(4)$	$+4$	$4$
$5$	Rectangle: $3 \times 4$	$+12$	$16$
$7$	Triangle: $\frac{1}{2}(2)(4)$	$+4$	$20$
$9$	Triangle: $\frac{1}{2}(2)(-2)$	$-2$	$18$

Analysis of $g$ :

Property	Answer	Reasoning
$g$ increasing	$(0, 7)$	$f > 0$ on $(0,7)$
$g$ decreasing	$(7, 9)$	$f < 0$ on $(7,9)$
Absolute max of $g$	$x = 7$ , $g(7)=20$	$f$ changes from $+$ to $-$
$g$ concave up	$(0, 2)$	$f$ increasing (slope $> 0$ )
$g$ concave down	$(2, 5)$ ? No, $f$ constant on $(2,5)$	$f'=0$ , so $g$ is linear there
Inflection points	$x=2, 5, 7$	$f$ changes from increasing to constant, etc.

Graph Analysis Practice 🎯

Combined Topics 🎯

Identify the correct analysis. 🔍

Compute from a graph. ✍️

Key Takeaways — Part 6

To find...	Look at...
$g(c)$	Signed area of $f$ from $a$ to $c$
$g$ increasing/decreasing	Sign of $f$
Local extrema of $g$	Sign changes of $f$
Concavity of $g$	Increasing/decreasing behavior of $f$
Inflection points of $g$	Local extrema of $f$

Up Next: Part 7 — Comprehensive Assessment.

Part 7: Comprehensive Assessment

Accumulation Functions

Part 7 of 7 — Comprehensive Assessment

Complete Formula Reference

Formula	Expression
Accumulation function	$g(x) = \int_a^x f(t)\,dt$
FTC Part 1	$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$
FTC + Chain Rule	$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))g'(x)$
Net change	$\int_a^b f'(t)\,dt = f(b)-f(a)$
Displacement	$\int_a^b v(t)\,dt$
Total distance	$\int_a^b
Average value	$\frac{1}{b-a}\int_a^b f(x)\,dx$
MVT for Integrals	$\exists\,c: f(c) = f_{avg}$

Top AP Mistakes

Mistake	Correction
Forgetting chain rule in FTC	$\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$
Confusing displacement and distance	Distance uses $
Wrong sign for lower-limit variable	$\frac{d}{dx}\int_x^b = -f(x)$
Forgetting $\frac{1}{b-a}$ in average value	Average = $\frac{\text{integral}}{\text{interval length}}$
Assuming $g(a) \ne 0$	$g(a) = \int_a^a f(t)\,dt = 0$ always
Reading $g$ from $f$ graph backwards	$g' = f$ , not $g = f'$

Quiz — FTC & Chain Rule 🎯

Quiz — Net Change & Average Value 🎯

Classify each scenario. 🔍

Final Challenge ✍️

Accumulation Functions — Complete!

You've mastered:

Part	Topic
1	Accumulation function definition & FTC Part 1
2	Graph interpretation & area computation
3	FTC with chain rule — all variations
4	Net change, displacement, rate in/out
5	Average value & MVT for integrals
6	Problem-solving workshop
7	Comprehensive assessment

You're ready for AP-level accumulation function problems!

← Back to Standard Lesson

Accumulation Functions

Accumulation Functions - Complete Interactive Lesson

Part 1: The Accumulation Concept

Accumulation Functions

Table of Contents

What is an Accumulation Function?

Key Properties at a Glance

Worked Example

Key Takeaways — Part 1

Part 2: Reading Graphs of f to Analyze F

Accumulation Functions

The Most Tested AP Skill

Complete Translation Guide

Computing F(x)F(x)F(x) from Geometric Areas

Worked Example

Key Takeaways — Part 2

Part 3: FTC Part 1 with Chain Rule Review

Accumulation Functions

Standard FTC Part 1

Chain Rule Extension

All Variations

Worked Examples

Key Takeaways — Part 3

Part 4: Net Change Applications

Accumulation Functions

The Net Change Theorem

Displacement vs. Total Distance

Rate In / Rate Out Problems

Key Takeaways — Part 4

Part 5: Average Value of a Function

Accumulation Functions

Average Value Formula

Mean Value Theorem for Integrals

Worked Example

Common Variations on AP Exams

Key Distinction

Key Takeaways — Part 5

Part 6: Practice Workshop

Accumulation Functions

Graph-to-Accumulation Strategy

Worked Example: Piecewise Linear Graph

Key Takeaways — Part 6

Part 7: Comprehensive Assessment

Accumulation Functions

Complete Formula Reference

Top AP Mistakes

Accumulation Functions — Complete!

Computing $F(x)$ from Geometric Areas