Unit 6: Integration and Accumulation of Change

Accumulation and the Meaning of the Definite Integral

What “accumulation” means (and why calculus cares)

In many real situations, you don’t directly measure a total amount; you measure a rate that tells you how the total is changing. For example, velocity tells you how position changes, flow rate (gallons per minute) tells you how much volume is added to a tank, and power usage rate (kilowatts) tells you how much energy is consumed over time. Accumulation is the idea of rebuilding the total change from the rate of change by adding up many small contributions over an interval.

Up to this point, derivatives have been the main tool for understanding rate of change (change per unit). Integrals reverse that perspective: an integral can represent the total change that builds up from a rate.

Signed area and “net change”

A key interpretation is geometric. If you graph a function f(x), then the definite integral

\int_a^b f(x)\,dx

represents the net (signed) area between the curve and the x-axis from x=a to x=b.

Where f(x) > 0, the integral contributes positive area; where f(x) < 0, it contributes negative area. This is why the definite integral is often described as net change.

A common confusion is mixing up net area with total area. Total area counts area below the axis as positive and is often written with absolute value:

\int_a^b |f(x)|\,dx

AP problems frequently test whether you understand that

\int_a^b v(t)\,dt

is displacement (net change in position), not total distance, while total distance traveled is

\int_a^b |v(t)|\,dt

Units: the quickest way to sanity-check an integral

Units make the meaning of an integral much clearer. If f(x) has units “something per x” (like meters per second or dollars per year), then

\int_a^b f(x)\,dx

has units “something,” because you multiply by dx. Example: if v(t) is in meters per second, then

\int_0^{10} v(t)\,dt

is in meters. This units check is one of the best ways to avoid interpreting an integral incorrectly on FRQs.

Definite integral vs indefinite integral (number vs family of functions)

A definite (bounded) integral has limits of integration and outputs a single number (a net accumulated amount):

\int_2^5 f(x)\,dx

An indefinite integral is also called an antiderivative and represents a whole family of functions:

\int f(x)\,dx

The bounds a and b matter: changing them changes the interval of accumulation.

Example 1: Net change from a rate

Suppose a particle has velocity (in meters per second)

v(t) = 3t - 4

Find the displacement from t=1 to t=4.

Displacement is the integral of velocity:

\int_1^4 (3t-4)\,dt

An antiderivative of 3t-4 is

\frac{3}{2}t^2 - 4t

\int_1^4 (3t-4)\,dt = \left[\frac{3}{2}t^2 - 4t\right]_1^4

Evaluate:

\left(\frac{3}{2}\cdot 16 - 16\right) - \left(\frac{3}{2}\cdot 1 - 4\right)

=(24-16) - (1.5-4)

=8 - (-2.5)=10.5

So the displacement is 10.5 meters.

A common pitfall is forgetting that if part of the interval has negative velocity, displacement can be smaller than total distance.

Exam Focus

Typical question patterns:
- Interpret \int_a^b r(t)\,dt as the net change in a quantity given a rate r(t).
- Distinguish displacement from total distance using \int |v(t)|\,dt.
- Use units to interpret what an integral represents.
Common mistakes:
- Treating a negative region as positive area when the question asks for net change.
- Forgetting that a definite integral is a number (and trying to attach “+C”).
- Ignoring units and misinterpreting the result.

Riemann Sums: Building the Definite Integral from Rectangles

Why Riemann sums exist

If a definite integral represents area/accumulation, how do you find it when the region has a curved boundary or you don’t have a nice formula? You approximate the region using shapes you do know. The standard starting point is to split the interval into rectangles, compute each rectangle’s area (base times height), and add them up. The more rectangles you use, the better the estimate. This method is a Riemann sum.

Conceptually:

Split the interval into subintervals.
Approximate the curve on each subinterval using a rectangle (or a trapezoid).
Add the approximations.
Make the subintervals thinner and thinner.

That limiting process leads to the definition of the definite integral.

Partitions, subinterval width, and sample points

A partition of [a,b] breaks the interval into n subintervals.

For equal-width subintervals:

\Delta x = \frac{b-a}{n}

The endpoints are

x_0 = a

x_1 = a+\Delta x

x_n = b

A Riemann sum uses a sample point x_i^* in each subinterval [x_{i-1},x_i], producing

\sum_{i=1}^n f(x_i^*)\Delta x

As n\to\infty (so \Delta x\to 0), if the limit exists, it becomes the definite integral:

\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x

Left, right, and midpoint sums (how AP usually frames them)

AP frequently specifies which sample points to use:

Left Riemann sum (use the left endpoints):

\sum_{i=1}^n f(x_{i-1})\Delta x

Right Riemann sum (use the right endpoints):

\sum_{i=1}^n f(x_i)\Delta x

Midpoint Riemann sum (use the midpoint of each subinterval):

x_i^* = \frac{x_{i-1}+x_i}{2}

\sum_{i=1}^n f\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x

A practical detail that shows up constantly with tables: for a left sum you do not use the furthest-right function value, and for a right sum you do not use the furthest-left function value.

The trapezoidal rule (a straighter-edge approximation)

Instead of rectangles, you can approximate each slice with a trapezoid whose top edge is the line segment connecting endpoint function values.

Geometrically, each trapezoid has area

\frac{1}{2}(b_1+b_2)h

where b_1 and b_2 are the two vertical “heights” (the function values at the endpoints) and h is the horizontal width of the subinterval.

For equal subintervals, the trapezoidal rule is

T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]

You can also view this as the average of the left and right sums:

T_n = \frac{L_n + R_n}{2}

Example of a single trapezoid computation (matching the standard formula): if a subinterval has width 1 and endpoint heights 2 and 5, then that trapezoid’s area is

\frac{1}{2}(2+5)(1)

From a table of values (even and uneven spacing)

A very common AP format gives a table of values of f at specific x values and asks you to approximate an integral using a left/right/midpoint/trapezoidal sum.

When the data are evenly spaced, \Delta x is the constant spacing.
When they’re not evenly spaced, you must use each subinterval’s actual width (for example, widths of 2, 2, and 3 in one table).

Example 1: Approximate with a left sum from a table

You are given values of f:

x	0	2	4	6
f(x)	5	4	1	-2

Approximate \int_0^6 f(x)\,dx using a left Riemann sum with the partition points in the table.

The subintervals are [0,2], [2,4], [4,6], each with width \Delta x = 2, and the left endpoints are 0, 2, 4.

L = f(0)\cdot 2 + f(2)\cdot 2 + f(4)\cdot 2

= 5\cdot 2 + 4\cdot 2 + 1\cdot 2 = 20

\int_0^6 f(x)\,dx \approx 20

Interpretation reminder: the negative value at x=6 didn’t matter for the left sum because x=6 is not used as a left endpoint here.

Example 2: Trapezoidal rule from the same table

Using

T = \frac{\Delta x}{2}\left[f(0)+2f(2)+2f(4)+f(6)\right]

with \Delta x=2:

T = 1\cdot \left[5+2(4)+2(1)+(-2)\right]

= 13

Notice how much smaller it is than the left sum because the function drops and becomes negative by the end.

Example 3: Table with uneven subinterval widths (left, right, midpoint idea, trapezoids)

Consider the tabular data:

x	0	2	4	7
f(x)	1	6	10	15

The widths are 2, 2, and 3.

A left sum using these subintervals is

(2)(1) + (2)(6) + (3)(10)

A right sum is

(2)(6) + (2)(10) + (3)(15)

A trapezoidal sum is

\frac{1}{2}(1+6)(2) + \frac{1}{2}(6+10)(2) + \frac{1}{2}(10+15)(3)

A midpoint rule requires function values at midpoints; if you approximate just the interval [0,4] as one subinterval, then the midpoint is 2, width is 4, and the midpoint rectangle estimate is

(4)(6)

This illustrates the core midpoint idea: use a rectangle whose height comes from the “in-between” value, not an endpoint.

Example 4: Tabular Riemann sums in context (and you often don’t need to simplify)

The AP exam frequently provides tables like this:

Years t	2	3	5	7	10
Height H(t)	1.5	2	6	11	15

A trapezoidal approximation over the given intervals is

\frac{1}{2}(1.5+2)(1) + \frac{1}{2}(2+6)(2) + \frac{1}{2}(6+11)(2) + \frac{1}{2}(11+15)(3)

A left sum is

(1)(1.5) + (2)(2) + (2)(6) + (3)(11)

A right sum is

(1)(2) + (2)(6) + (2)(11) + (3)(15)

You do not always have to simplify these expressions unless the question explicitly requires a numerical value.

Exam Focus

Typical question patterns:
- Approximate a definite integral from a table using left/right/midpoint sums or the trapezoidal rule.
- Write a Riemann sum expression that represents an integral (or identify the integral represented by a sum).
- Decide whether an approximation is an overestimate or underestimate using increasing/decreasing or concavity.
Common mistakes:
- Using the wrong endpoints (left vs right) or accidentally including one too many terms.
- Forgetting to multiply by the interval width (using values but not \Delta x).
- Assuming equal spacing when the table spacing is not uniform.

The Fundamental Theorem of Calculus (FTC): Connecting Derivatives and Integrals

Why the FTC is the centerpiece of integration

Riemann sums define integrals, but they’re not practical for exact computation. The Fundamental Theorem of Calculus explains that integration and differentiation are inverse processes in a precise way. This gives you a powerful shortcut: instead of taking a limit of sums, you evaluate antiderivatives.

FTC Part 1: Derivative of an accumulation function

Define an accumulation function

F(x) = \int_a^x f(t)\,dt

If f is continuous, FTC Part 1 states:

F'(x) = f(x)

In plain language: if you accumulate the rate f(t) from a up to x, then the rate at which that accumulated total changes at x is just the original rate value at x.

When the upper limit is not just x

G(x) = \int_a^{g(x)} f(t)\,dt

then

G'(x) = f(g(x))g'(x)

If the variable is in the lower limit,

H(x) = \int_{g(x)}^a f(t)\,dt

then

H'(x) = -f(g(x))g'(x)

Example 1: Differentiate an accumulation function

Let

G(x) = \int_2^{x^3} \sqrt{1+t^2}\,dt

Then

G'(x) = \sqrt{1+(x^3)^2}\cdot 3x^2

G'(x) = 3x^2\sqrt{1+x^6}

A common pitfall is “differentiating inside” with respect to x. FTC Part 1 says to keep the integrand as a function of the dummy variable, plug in the limit, and multiply by the derivative of the limit.

FTC Part 2: Evaluate a definite integral using an antiderivative

If F'(x)=f(x), then

\int_a^b f(x)\,dx = F(b) - F(a)

Equivalently,

\left[F(x)\right]_a^b = F(b)-F(a)

This is the rule that turns “bounded integrals” into plug-in-and-subtract computations.

Net Change Theorem (a conceptual restatement)

If a quantity has rate of change F'(x), then

F(b) - F(a) = \int_a^b F'(x)\,dx

Example 2: Use FTC Part 2 to evaluate an integral

Evaluate

\int_1^3 (2x + 5)\,dx

An antiderivative is

x^2 + 5x

\left[x^2 + 5x\right]_1^3

= (9+15) - (1+5) = 18

A common pitfall is adding +C to a definite integral evaluation. Constants cancel in F(b)-F(a), so you do not include them.

Example 3: Definite integral of a simple power rule antiderivative

Compute

\int_2^3 2x\,dx

An antiderivative of 2x is

x^2

So the definite integral is

\left[x^2\right]_2^3

= 3^2 - 2^2 = 5

Exam Focus

Typical question patterns:
- Differentiate functions defined by integrals (especially with g(x) as a limit).
- Evaluate definite integrals by finding antiderivatives (often mixed with properties).
- Use the Net Change Theorem to connect a derivative/rate to total change.
Common mistakes:
- Forgetting the chain rule factor g'(x) when the bound is g(x).
- Treating \int_a^x f(t)\,dt like an ordinary product instead of a function definition.
- Writing +C for a definite integral or failing to plug in both bounds.

Properties of Definite Integrals (How to Simplify Without Integrating)

Why properties matter

Not every integral on an AP exam is meant to be computed by finding an antiderivative. Sometimes the integrand is messy, or you’re only given a graph or a few key values. Properties of definite integrals let you rewrite, break apart, or compare integrals to find exact values efficiently.

Linearity: add and scale

If f and g are integrable and c is a constant, then

\int_a^b (f(x)+g(x))\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx

and

\int_a^b c f(x)\,dx = c\int_a^b f(x)\,dx

Additivity over intervals

If a < c < b, then

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

Reversing bounds changes the sign

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx

A special case:

\int_a^a f(x)\,dx = 0

Symmetry: even and odd functions

If f is even, meaning f(-x)=f(x), then

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx

If f is odd, meaning f(-x)=-f(x), then

\int_{-a}^{a} f(x)\,dx = 0

These shortcuts only apply on symmetric intervals of the form [-a,a].

Comparison ideas (often used with graphs)

If f(x) \ge g(x) on [a,b], then

\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx

If m \le f(x) \le M on [a,b], then

m(b-a) \le \int_a^b f(x)\,dx \le M(b-a)

Example 1: Use properties to compute an integral from given values

Suppose you know:

\int_0^2 f(x)\,dx = 5

and

\int_2^6 f(x)\,dx = -1

Find

\int_6^0 f(x)\,dx

First,

\int_0^6 f(x)\,dx = 5 + (-1) = 4

Then reverse bounds:

\int_6^0 f(x)\,dx = -\int_0^6 f(x)\,dx = -4

Example 2: Symmetry shortcut

Let f be odd. Evaluate

\int_{-3}^{3} (f(x) + 2)\,dx

Split using linearity:

\int_{-3}^{3} f(x)\,dx + \int_{-3}^{3} 2\,dx

Odd symmetry gives

\int_{-3}^{3} f(x)\,dx = 0

And

\int_{-3}^{3} 2\,dx = 2(3-(-3)) = 12

So the value is 12.

Exam Focus

Typical question patterns:
- Compute an integral using only given integral values and properties (no antiderivative needed).
- Use symmetry (even/odd) on symmetric intervals to simplify.
- Rewrite integrals by splitting intervals or reversing bounds.
Common mistakes:
- Forgetting the negative sign when reversing bounds.
- Misapplying even/odd rules on non-symmetric intervals.
- Distributing incorrectly: thinking \int (f+g) = \int f\cdot \int g (this is false).

Antiderivatives and Indefinite Integrals (Recovering Functions from Their Derivatives)

What an antiderivative is (and why +C matters)

An antiderivative of f(x) is a function F(x) such that

F'(x) = f(x)

Because derivatives “forget” constants (the derivative of any constant is 0), antiderivatives come in families. If F(x) is one antiderivative, then F(x)+C is also an antiderivative for any constant C. That’s why the constant of integration is very important.

Indefinite integral notation

The indefinite integral is written

\int f(x)\,dx = F(x) + C

Important distinction:

A definite integral is a number.
An indefinite integral is a family of functions.

Core antiderivative patterns (reverse-derivative thinking)

Antidifferentiation is often the “opposite” of differentiation. For the power rule, if differentiating multiplies down and decreases the power, antidifferentiating divides and increases the power.

Power rule (for n \ne -1):

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C

This includes common simplifications like

\int 2x\,dx = x^2 + C

because

\int 2x\,dx = \frac{2x^2}{2} + C

If an integrand is not immediately in power-rule format, you often algebraically manipulate it so that it is.

Special logarithm case:

\int \frac{1}{x}\,dx = \ln|x| + C

Exponential:

\int e^x\,dx = e^x + C

\int a^x\,dx = \frac{a^x}{\ln(a)} + C

Trigonometric basics (it’s efficient to memorize derivative-antiderivative pairs for speed on the AP exam):

\int \cos x\,dx = \sin x + C

\int \sin x\,dx = -\cos x + C

For example, since

\frac{d}{dx}(\sin x) = \cos x

it follows that

\int \cos x\,dx = \sin x + C

Example 1: Find a function from its derivative

Given

f'(x) = 6x^2 - \frac{4}{x}

and f(1)=3, find f(x).

Integrate:

f(x) = \int \left(6x^2 - \frac{4}{x}\right)dx

Compute each part:

\int 6x^2\,dx = 2x^3

\int \frac{4}{x}\,dx = 4\ln|x|

f(x) = 2x^3 - 4\ln|x| + C

Use f(1)=3:

3 = 2(1)^3 - 4\ln|1| + C

Since \ln|1|=0,

3 = 2 + C

So C=1 and

f(x) = 2x^3 - 4\ln|x| + 1

A common pitfall is dropping the absolute value in \ln|x|.

Example 2: Indefinite integral using linearity and patterns

Evaluate

\int (5e^x - 3\cos x)\,dx

Using linearity:

\int 5e^x\,dx - \int 3\cos x\,dx

5e^x - 3\sin x + C

Exam Focus

Typical question patterns:
- Find f(x) given f'(x) and an initial condition.
- Compute indefinite integrals using basic patterns and linearity.
- Recognize when \ln|x| appears.
Common mistakes:
- Forgetting +C on an indefinite integral.
- Using the power rule when n=-1 (you must use \ln|x|).
- Losing track of signs on trig antiderivatives, especially \int \sin x\,dx.

Integration by Substitution (Reverse Chain Rule)

Why substitution is needed

Basic antiderivative rules work when the integrand matches a known pattern. But many functions are composites, like \sin(3x) or a product like \sqrt{1+x^2}\cdot 2x. Differentiation of composites uses the chain rule, so integration of composites often requires reversing the chain rule.

Integration by substitution (u-substitution) is the main method in this unit for integrals that involve a function and (up to a constant factor) its derivative.

The core idea: spot an inner function and its derivative

If you see

f(g(x))g'(x)

then you can set

u=g(x)

and rewrite the integral in terms of u:

\int f(g(x))g'(x)\,dx = \int f(u)\,du

A reliable u-substitution process

Choose u to be an “inside” expression (often inside parentheses, a root, an exponent, or a trig function).
Differentiate: du = u'(x)dx.
Rewrite the integral entirely in terms of u and du.
Integrate with respect to u.
Substitute back to x.

U-substitution can be tricky at first, but it’s extremely helpful once you learn to recognize patterns.

Example 1: Indefinite integral (classic pattern)

Evaluate

\int 2x\cos(x^2)\,dx

Let

u = x^2

Then

du = 2x\,dx

\int 2x\cos(x^2)\,dx = \int \cos(u)\,du

Integrate:

\sin(u) + C

Substitute back:

\sin(x^2) + C

Example 2: Definite integral with substitution and changed bounds

Evaluate

\int_1^2 \frac{x}{1+x^2}\,dx

Let

u = 1+x^2

Then

du = 2x\,dx

x\,dx = \frac{1}{2}du

Change bounds: when x=1, u=2; when x=2, u=5.

Rewrite:

\int_1^2 \frac{x}{1+x^2}\,dx = \int_2^5 \frac{1}{u}\cdot \frac{1}{2}\,du

\frac{1}{2}\int_2^5 \frac{1}{u}\,du

Evaluate:

\frac{1}{2}\left[\ln|u|\right]_2^5

\frac{1}{2}(\ln 5 - \ln 2)

Equivalently,

\frac{1}{2}\ln\left(\frac{5}{2}\right)

Example 3: U-sub with a shifted power

Evaluate

\int (x - 4)^{10}\,dx

Let

u = x-4

Then

\frac{du}{dx} = 1

dx = du

Rewrite and integrate:

\int u^{10}\,du

\frac{u^{11}}{11} + C

Substitute back:

\frac{(x-4)^{11}}{11} + C

Recognizing substitution patterns quickly

Substitution is likely when you see:

A composite like \sin(\text{something}) or e^{\text{something}} along with the derivative of “something.”
Rational forms like \frac{f'(x)}{f(x)}, which lead to logarithms.
Powers like (\text{something})^n with a matching derivative factor.

Exam Focus

Typical question patterns:
- Evaluate an integral using u-substitution (often set up so the derivative of the inner function is present).
- Evaluate a definite integral using substitution, sometimes with changed bounds.
- Recognize logarithm outcomes from \int f'(x)/f(x)\,dx.
Common mistakes:
- Missing a constant factor: having du=2x\,dx but only x\,dx appears (you must adjust with a factor).
- Changing bounds to u but then evaluating with x anyway.
- Picking a u that doesn’t simplify the integral.

Accumulation Functions and Interpreting Integrals in Context

Accumulation functions as total change up to a variable endpoint

A common AP theme is defining a new function using an integral, such as

A(x) = \int_a^x r(t)\,dt

This function represents the accumulated amount from a to x.

It links multiple interpretations:

Graphical: signed area under r(t) from a to x.
Contextual: net change in a quantity.
Analytical: by FTC Part 1, the derivative of the accumulation function is the original rate.

“Rate in, total out” and the role of an initial value

Many word problems provide a starting amount and a rate of change. If you know Q(a) and Q'(t)=r(t), then

Q(b) = Q(a) + \int_a^b r(t)\,dt

A frequent mistake is computing the integral (the change) and forgetting to add the initial value when the question asks for the final amount.

Displacement vs position (and similar pairs)

If s(t) is position, then s'(t)=v(t).

Displacement on [a,b]:

s(b)-s(a)=\int_a^b v(t)\,dt

Position function from velocity and initial position:

s(t)=s(a)+\int_a^t v(u)\,du

This pattern repeats in many contexts: temperature from a heating rate, population from a growth rate, mass from net inflow/outflow, and so on.

Example 1: Build a quantity from a rate and initial amount

A tank contains 50 gallons of water at t=0. Water flows in at rate

R(t)=6-0.5t

gallons per minute for 0\le t\le 8. How much water is in the tank at t=8 (assuming no outflow)?

Final amount equals initial plus accumulated inflow:

V(8)=50+\int_0^8 (6-0.5t)\,dt

An antiderivative is

6t-0.25t^2

\left[6t-0.25t^2\right]_0^8

=(48-16)-0 = 32

Then

V(8)=50+32=82

So the tank has 82 gallons at t=8.

Example 2: Using a graph conceptually (increasing/decreasing and concavity)

Suppose you’re shown a graph of a rate function r(t) and asked where

A(x)=\int_0^x r(t)\,dt

is increasing or decreasing. By FTC Part 1,

A'(x)=r(x)

So A(x) increases where r(x)>0, decreases where r(x)