Unit 6: Integration and Accumulation of Change

Accumulation and the Net Change Idea

A big shift happens when you move from derivatives to integrals. With derivatives, you ask: “At this instant, how fast is something changing?” With integrals, you ask: “Over an interval of time (or distance), how much change has accumulated?” In that sense, the integral is tightly connected to the idea of an antiderivative: instead of “change per unit,” you’re tracking “total change over many units.”

Rates create totals: the core story of integration

Suppose you know a rate of change function r(t). For example:

  • r(t) could be water flowing into a tank in liters per minute.
  • r(t) could be a car’s velocity in meters per second.
  • r(t) could be revenue earned per day.

If you want the total amount accumulated from time t=a to t=b, you might first think “multiply rate by time.” That works only if the rate is constant. If the rate varies, you break the interval into tiny pieces where the rate is “almost constant,” add up each tiny contribution, then take a limit. That limiting process becomes the definite integral.

Net change vs. total change

When a rate can be positive or negative, integration naturally tracks net change.

  • Net change counts positive contributions and subtracts negative contributions.
  • Total accumulated amount (like total distance traveled) often requires integrating the absolute value of the rate.

A classic example is velocity v(t):

  • Displacement (net change in position) from a to b is the integral of velocity.
  • Total distance traveled is the integral of speed |v(t)|.

It’s easy to confuse these. AP problems often test whether you know when negative values should subtract.

The Net Change Theorem

If a quantity Q(t) changes with rate Q'(t), then the net change in Q from a to b is

Q(b)-Q(a)=\int_a^b Q'(t)\,dt

This is less a “formula to memorize” and more a translation rule:

  • If you’re given a rate, integrating gives net change.
  • If you know an initial value, you can reconstruct the quantity later.

Example 1: Accumulation from a rate (conceptual)

A tank’s water volume V(t) changes at rate V'(t)=r(t) liters/min. If r(t) is sometimes negative (draining), then

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

is the net change in volume over the first 10 minutes. If the tank started with V(0)=50 liters, then

V(10)=50+\int_0^{10} r(t)\,dt

Notice how integration is acting like “accumulated change added onto the initial amount.”

Example 2: Displacement vs. distance

If a particle’s velocity is v(t) on [0,5], then

  • Displacement: \int_0^5 v(t)\,dt
  • Total distance: \int_0^5 |v(t)|\,dt

If the graph of v(t) dips below the axis, the area below counts as negative displacement, but distance still counts it as positive.

Exam Focus
  • Typical question patterns:
    • “A quantity changes at rate r(t); write an expression for the total (net) change on [a,b] and interpret it.”
    • “Given a rate graph/table, find accumulated change and then compute the quantity using an initial condition.”
    • “Distinguish between displacement and total distance using integrals.”
  • Common mistakes:
    • Treating negative rate values as still adding (confusing net change with total accumulation).
    • Forgetting to add the initial value when asked for the final amount.
    • Mixing up units: a rate times time should produce a quantity in the original units.

Approximating Accumulation with Riemann Sums

When you don’t have a simple formula (or you’re building the idea of the integral from scratch), you estimate “area under a curve” or “total change” using shapes you do know. The most basic approach is rectangles: split the region into rectangles, add their areas, and notice that more rectangles generally gives a better estimate. This method is called a Riemann sum.

Partitioning an interval

Take an interval [a,b] and split it into n subintervals. If the partition is equal-width, then each subinterval has width

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

You pick a representative point x_i^* in each subinterval (left endpoint, right endpoint, midpoint, or any point). Then you approximate the accumulated change by

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

Conceptually:

  • f(x_i^*) is “how tall” the function is on that tiny slice.
  • \Delta x is the “width” of the slice.
  • Multiplying gives a small rectangle-like area, and summing adds them up.

Left, right, and midpoint sums

Different choices of x_i^* produce common named approximations:

  • Left Riemann sum: use the left endpoint of each subinterval.
  • Right Riemann sum: use the right endpoint.
  • Midpoint Riemann sum: use the midpoint.

The rectangle formula idea is always “base times height,” meaning each piece looks like

\text{(width)}\cdot\text{(function value)}

If f is increasing on [a,b], then the left sum tends to **underestimate** and the right sum tends to **overestimate**. If f is decreasing, the roles reverse.

Writing a Riemann sum from a formula

If you partition [a,b] into n equal parts, the right endpoints are

x_i=a+i\Delta x

So a right Riemann sum is

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

A midpoint sum uses

x_i^*=a+\left(i-\frac{1}{2}\right)\Delta x

giving

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

On AP Free Response, a common skill is recognizing which expression matches which sum.

Example 1: Build a midpoint Riemann sum (structure)

Approximate \int_0^4 f(x)\,dx with n=4 subintervals using midpoints.

  1. Compute width:

\Delta x=\frac{4-0}{4}=1

  1. Midpoints of [0,1],[1,2],[2,3],[3,4] are 0.5,1.5,2.5,3.5.

  2. Midpoint sum:

f(0.5)(1)+f(1.5)(1)+f(2.5)(1)+f(3.5)(1)

Example 2: Riemann sum from a table (equal widths)

Suppose you’re given values of a rate function r(t) at times t=0,2,4,6 (minutes) and asked for a left sum on [0,6] using those subintervals.

  • The width is \Delta t=2 minutes.
  • Left endpoints are t=0,2,4.

Left sum:

r(0)(2)+r(2)(2)+r(4)(2)

Students often mistakenly include r(6) in a left sum; that would be a right sum.

Example 3: Riemann and trapezoidal sums from an unevenly spaced table

Often the AP gives a table where the widths are not all the same. Consider the data:

x0247
f(x)161015

The subinterval widths are 2,2,3.

  • Left sum (use left endpoints, so do not use the furthest right value):

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

  • Right sum (use right endpoints, so do not use the furthest left value):

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

  • Midpoint (shown in a “not complete, but notice the idea” style):

(4)(6)

This highlights that midpoint methods need function values at midpoints; if the table doesn’t provide them, you may not be able to do a true midpoint sum without extra information.

  • Trapezoid (sum of trapezoids on each subinterval):

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

Exam Focus
  • Typical question patterns:
    • “Write a Riemann sum expression that approximates \int_a^b f(x)\,dx using right endpoints and n subintervals.”
    • “Approximate an integral from a table using left/right/midpoint sums.”
    • “Determine whether a left or right sum overestimates/underestimates given monotonicity.”
  • Common mistakes:
    • Mixing up left vs. right endpoints, especially when given a table.
    • Using n subintervals but listing n+1 rectangles (or vice versa).
    • Forgetting that the width is \Delta x (or \Delta t), not necessarily 1.
    • Attempting a midpoint sum from a table that doesn’t include midpoint heights.

Definite Integrals as Limits of Riemann Sums

Riemann sums are approximations. The definite integral is what you get when the approximation becomes perfect in the limit.

From approximation to definition

If you let the number of subintervals n increase, the rectangles get thinner and the approximation improves. The definite integral is defined by the limit

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

when this limit exists.

What the integral represents (not just “area”)

It’s common to say “the definite integral is area under the curve.” That is fully true only when f(x)\ge 0.

A more accurate statement:

  • \int_a^b f(x)\,dx is the **signed area** between the curve and the x-axis.
  • Regions above the axis contribute positively.
  • Regions below contribute negatively.

In real contexts (rates), that signed behavior corresponds to net gain vs. net loss.

Notation and vocabulary

The notation

\int_a^b f(x)\,dx

has several pieces:

  • \int: integral sign (think “sum”).
  • a,b: bounds of integration.
  • f(x): integrand.
  • dx: indicates the variable of integration and the “width” direction.

The two numbers (bounds) mean the integral is definite or bounded, and you are accumulating from x=a to x=b.

Example: Evaluating an integral from geometry

If you are given the graph of f(x) and it forms simple shapes, you can compute the integral without any antiderivative formulas.

Suppose on [0,2] the graph is a line from (0,0) to (2,4), forming a triangle above the axis. Then

  • Base = 2
  • Height = 4
  • Area =

\frac{1}{2}(2)(4)=4

So

\int_0^2 f(x)\,dx=4

If the same triangle were below the axis, the integral would be -4.

Properties of integrals (built from “signed area”)

  1. Reversing bounds changes the sign:

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

  1. Zero width gives zero accumulation:

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

  1. Additivity across intervals:

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

  1. Constant multiple and sum rules:

\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx

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

  1. Comparison: if f(x)\ge g(x) on [a,b] then

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

Exam Focus
  • Typical question patterns:
    • “Use the definition of the integral as a limit of Riemann sums to identify the integral represented by a sum.”
    • “Given areas of regions on a graph, compute integrals using sign.”
    • “Use properties (additivity, reversal, linearity) to simplify an expression of integrals.”
  • Common mistakes:
    • Treating all integrals as positive “area” even when the graph is below the axis.
    • Dropping the negative sign when bounds are reversed.
    • Misusing additivity (for example, forgetting that splitting at c requires matching bounds exactly).

The Fundamental Theorem of Calculus (FTC) and Antiderivatives

The Fundamental Theorem of Calculus is the bridge between two worlds: the derivative world (instantaneous rate) and the accumulation world (integrals). It explains why antidifferentiation can compute definite integrals and how to differentiate certain integral expressions.

FTC Part 1: Derivative of an accumulation function

Define an accumulation function

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

If f is continuous, then

A'(x)=f(x)

The variable inside the integral is a dummy variable, so writing f(t) helps you see that the upper limit is the true input.

Chain rule versions (very common on AP)

If the upper limit is g(x):

\frac{d}{dx}\left(\int_a^{g(x)} f(t)\,dt\right)=f(g(x))g'(x)

If both bounds depend on x:

\frac{d}{dx}\left(\int_{h(x)}^{g(x)} f(t)\,dt\right)=f(g(x))g'(x)-f(h(x))h'(x)

Example 1: Differentiate an integral with a variable upper limit

Let

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

Then

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

So

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

Example 2: Both bounds vary

Let

H(x)=\int_{\sin x}^{x} (1+t^4)\,dt

Then

H'(x)=(1+x^4)\cdot 1-(1+(\sin x)^4)\cdot \cos x

So

H'(x)=1+x^4-\cos x\left(1+\sin^4 x\right)

FTC Part 2: Evaluating a definite integral using an antiderivative

If F'(x)=f(x) on an interval, then

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

This is the rule many courses informally call the “first fundamental theorem” for evaluating bounded integrals: find an antiderivative, plug in the top bound, plug in the bottom bound, and subtract.

Example 3: Evaluate a definite integral (basic)

Evaluate

\int_1^3 (2x+1)\,dx

An antiderivative is

F(x)=x^2+x

Then

\int_1^3 (2x+1)\,dx=F(3)-F(1)

Compute

F(3)=12

F(1)=2

So

\int_1^3 (2x+1)\,dx=10

Example 4: A definite integral of 2x from 2 to 3

Since an antiderivative of 2x is x^2,

\int_2^3 2x\,dx=3^2-2^2

This is a concrete example of “plug in the top, minus plug in the bottom.”

Example 5: Using FTC with a known derivative relationship

If you are told F'(x)=\cos(x^2) and asked for

\int_0^2 \cos(x^2)\,dx

you might not be able to find an elementary antiderivative directly, but if F is given (as a function, table, or graph), then

\int_0^2 \cos(x^2)\,dx=F(2)-F(0)

Indefinite integrals, the power rule, and the constant of integration

An indefinite integral represents a family of antiderivatives. The constant of integration is crucial because differentiating any constant gives zero, so reversing differentiation cannot recover which constant was there.

For power functions (the most common AB skill), the reverse of the derivative power rule is:

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

for n\ne -1.

For example, the integral of 2x can be written as

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

which simplifies to

x^2+C

If an integrand is not obviously in power-rule format, algebraically manipulate it until it is (for instance, rewrite, expand, or factor as needed).

“Advanced” (still AB) integration tools: trig antiderivatives and substitution

Some integrals are hard to force into a power rule directly. Two common approaches are:

1) Trig antiderivatives: since

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

the corresponding integral is

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

On a timed exam, it’s efficient to memorize basic derivative-trig pairs so you can reverse them quickly.

2) U-substitution (substitution):

  • Choose a term to be your u.
  • Differentiate to get \frac{du}{dx}.
  • Substitute: replace the chosen expression with u and replace dx using du.
  • Integrate in terms of u.

Example:

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

Let

u=x-4

Then

\frac{du}{dx}=1

So

dx=\frac{du}{1}

Rewrite and integrate:

\int u^{10}\,du=\frac{u^{11}}{11}+C

Substitute back:

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

Exam Focus
  • Typical question patterns:
    • “Differentiate an expression like \int_a^{g(x)} f(t)\,dt.”
    • “Evaluate a definite integral using an antiderivative and FTC.”
    • “Given that F' = f, compute \int_a^b f(x)\,dx from values of F.”
    • “Find an indefinite integral and include +C.”
    • “Recognize when a basic trig antiderivative or a simple substitution is the fastest method.”
  • Common mistakes:
    • Forgetting the chain rule factor g'(x) when the limit is g(x).
    • Using f(x) instead of f(g(x)).
    • Treating the lower bound’s contribution as positive (missing the subtraction term).
    • Forgetting +C on an indefinite integral.

Accumulation Functions and Interpreting Their Graphs

An accumulation function is a function defined by an integral with a variable limit. These show up constantly because they model “total so far.”

Defining accumulation functions

A typical form is

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

Key interpretations:

  • A(x) measures the net signed area under f from a to x.
  • A(a)=0 because you have accumulated nothing over a zero-length interval.
  • The rate of change of A is the current value of f:

A'(x)=f(x)

So:

  • Where f(x)>0, A(x) is increasing.
  • Where f(x)

How concavity of an accumulation function works

Differentiate again:

A''(x)=f'(x)

So:

  • If f is increasing, then A is concave up.
  • If f is decreasing, then A is concave down.

Concavity depends on whether f is increasing or decreasing, not on whether f is above or below the axis.

Example 1: Where is the accumulation function increasing?

Let

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

If a graph shows f(t) is positive on [0,2] and negative on [2,5], then:

  • A increases on [0,2].
  • A decreases on [2,5].

At x=2, f(2)=0, so A'(2)=0 and A has a critical point.

Example 2: Identify a maximum of an accumulation function

Since A'(x)=f(x), A has a local maximum where f changes from positive to negative. On a graph of f, look for an x-intercept where the sign switches from above to below.

Accumulation with non-constant lower limit

Sometimes you’ll see

B(x)=\int_x^a f(t)\,dt

Using properties of integrals,

\int_x^a f(t)\,dt=-\int_a^x f(t)\,dt

So if A(x)=\int_a^x f(t)\,dt then B(x)=-A(x), and

B'(x)=-f(x)

Exam Focus
  • Typical question patterns:
    • “Given the graph of f, determine where A(x)=\int_a^x f(t)\,dt is increasing/decreasing or has extrema.”
    • “Use A'(x)=f(x) and A''(x)=f'(x) to deduce slope and concavity of an accumulation function.”
    • “Differentiate accumulation functions with variable bounds (including lower bound as x).”
  • Common mistakes:
    • Confusing the sign of A with whether it’s increasing (increasing depends on f, not on A).
    • Mixing up concavity rules (concavity depends on f', not on whether f is positive).
    • Missing a negative sign when x is in the lower bound.

Numerical Integration: Trapezoidal Rule and Reading Data

In many realistic situations, you do not have a simple formula for f(x). You might have a table of measured values or a complicated expression whose antiderivative is not accessible. Then you approximate the integral numerically.

Why trapezoids improve on rectangles

Riemann sums approximate the curve by flat rectangles. The Trapezoidal Rule approximates the curve by straight-line segments between data points, creating trapezoids. This often reduces error when the function is reasonably smooth.

A single trapezoid area idea is

\text{Area of trapezoid}=\frac{1}{2}(b_1+b_2)h

where the “bases” are the left and right heights and the “height” is the subinterval width.

For example, if one subinterval has width 1 and endpoint heights 2 and 5, that trapezoid’s area is

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

The Trapezoidal Rule (equal subintervals)

If you partition [a,b] into n equal subintervals of width \Delta x and have values f(x_0), f(x_1), \dots, f(x_n), then

\int_a^b f(x)\,dx\approx\frac{\Delta x}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1})+f(x_n)\right)

Equivalently, it’s the sum of trapezoids

\frac{\Delta x}{2}(f(x_{i-1})+f(x_i))

Example 1: Trapezoidal Rule from a table

You are given the table of f(x) at x=0,1,2,3:

x0123
f(x)2541

Approximate \int_0^3 f(x)\,dx.

Here \Delta x=1 and n=3.

\int_0^3 f(x)\,dx\approx\frac{1}{2}\left(f(0)+2f(1)+2f(2)+f(3)\right)

Substitute values:

\frac{1}{2}\left(2+2\cdot 5+2\cdot 4+1\right)

Compute:

\frac{1}{2}(21)=10.5

Example 2: Left vs. right vs. trapezoid with the same data

With the same table and \Delta x=1:

  • Left sum:

f(0)+f(1)+f(2)=11

  • Right sum:

f(1)+f(2)+f(3)=10

  • Trapezoid (average of left and right when widths are equal):

\frac{11+10}{2}=10.5

Tabular trapezoids, left sums, and right sums (you don’t have to simplify)

The majority of the time when you have to use a Riemann Sum, the AP gives it to you in tabular format. For example:

Years: t235710
Height: H(t)1.5261115

The subinterval widths are 1,2,2,3.

  • Trapezoids:

\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)

  • Left sum:

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

  • Right sum:

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

You do not have to simplify these unless the problem explicitly asks for a simplified numerical value.

Error: what you should and should not claim

You are often asked to compute approximations but not full error bounds. You should reason qualitatively:

  • If f is concave up, trapezoids tend to overestimate.
  • If f is concave down, trapezoids tend to underestimate.
Exam Focus
  • Typical question patterns:
    • “Approximate \int_a^b f(x)\,dx using the trapezoidal rule given a table.”
    • “Compare left, right, midpoint, and trapezoidal approximations and reason about over/underestimates using increasing/decreasing or concavity.”
    • “Use numerical integration to estimate a physical quantity (total flow, total change) from data.”
  • Common mistakes:
    • Using the wrong \Delta x when x-values are not 1 unit apart.
    • Doubling endpoints in the trapezoidal formula.
    • Forgetting to double the interior values in the equal-width trapezoid formula.
    • Claiming an estimate is exact just because it uses trapezoids.

Interpreting Definite Integrals in Context (Units and Meaning)

AP questions often test whether you can explain what an integral represents in words and whether you can correctly interpret its units.

Units: integrals multiply units

If f(x) has units “something per x-unit,” then integrating over x multiplies by the x-unit.

  • If f(t) is liters per minute and you integrate in minutes, the result is liters.
  • If v(t) is meters per second and you integrate in seconds, the result is meters.

More generally, if f(x) has units U and x has units X, then \int f(x)\,dx has units U\cdot X.

The integral as “accumulated quantity”

To interpret

\int_a^b r(t)\,dt

you say: “the net amount accumulated from t=a to t=b when the rate of change is r(t).”

Signed area interpretation vs. physical meaning

If the integrand represents a physical rate that can be negative (like velocity or net flow), then negative values correspond to the quantity decreasing. If the situation demands a nonnegative physical total (like total distance), you should integrate |v(t)| instead.

Example 1: Interpreting an integral with words and units

A bakery’s production rate is p(t) cakes/hour. Then

\int_2^6 p(t)\,dt

represents the total number of cakes produced from hour 2 to hour 6. Units work out as cakes.

Example 2: Net change and initial value

If a population N(t) changes at rate N'(t)=r(t) people/year and N(0)=1200, then

N(5)=1200+\int_0^5 r(t)\,dt

A common mistake is writing N(5)=\int_0^5 r(t)\,dt and ignoring the initial population.

Exam Focus
  • Typical question patterns:
    • “Interpret \int_a^b f(x)\,dx in context (what does it mean?). Include units.”
    • “Given a rate and an initial value, compute the final value and explain reasoning.”
    • “Distinguish when an integral gives net change vs. when you need total accumulation (absolute value).”
  • Common mistakes:
    • Incorrect or missing units in explanations.
    • Interpreting any integral as “area” rather than “net change” in a rate context.
    • Forgetting that negative rate values reduce the accumulated total.

Connecting Integrals and Derivatives Through Motion (A Key Application)

Motion problems are one of the clearest places where Unit 6 ideas come together: rates, accumulation, and interpretation.

Velocity, position, and acceleration relationships

If s(t) is position, then

  • Velocity: v(t)=s'(t)
  • Acceleration: a(t)=v'(t)=s''(t)

The integral relationships reverse these derivatives:

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

and

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

Displacement from velocity

Displacement is the signed area under the velocity graph. If an object moves forward then backward, net displacement may be small even if the object traveled far.

Total distance traveled

Total distance is accumulation of speed, not velocity:

\text{Total distance}=\int_a^b |v(t)|\,dt

Example 1: Displacement and position from a velocity graph

Suppose the graph of v(t) shows:

  • On [0,2], the area above the axis is 6.
  • On [2,5], the area below the axis has magnitude 4.

Then displacement on [0,5] is

\int_0^5 v(t)\,dt=6-4=2

If s(0)=10, then

s(5)=10+2=12

Total distance traveled is 6+4=10.

Example 2: Change in velocity from acceleration data

If acceleration is measured at discrete times and you approximate \int_a^b a(t)\,dt with a trapezoidal rule, that result is the change in velocity, v(b)-v(a). A common error is interpreting \int a(t)\,dt as velocity itself rather than change in velocity.

Exam Focus
  • Typical question patterns:
    • “Given v(t) (graph/table), compute displacement, total distance, or position with an initial condition.”
    • “Given a(t) (graph/table), compute change in velocity and then velocity at a time.”
    • “Interpret the meaning of an integral in motion language.”
  • Common mistakes:
    • Reporting displacement when the question asks for total distance (forgetting absolute value).
    • Forgetting to add initial position or initial velocity.
    • Confusing velocity with acceleration when interpreting integrals.

Putting It All Together: Multi-Representation Reasoning with Integrals

Unit 6 problems often combine formulas, graphs, and tables. The key skill is translating between them using the meaning of the integral.

From a graph: compute integrals using geometry and sign

When you have a graph but no formula, look for shapes (rectangles, triangles, trapezoids, semicircles if clearly indicated). Treat areas below the axis as negative for the integral.

From a table: choose an appropriate numerical method

If values are given at evenly spaced points:

  • Left and right sums use endpoint values directly.
  • Midpoint sums may require intermediate points (sometimes provided).
  • Trapezoidal rule is common and often expected.

If spacing is not uniform, you can still apply “sum of trapezoids” using each subinterval’s width individually. For subinterval [x_{i-1},x_i],

\text{Area}\approx\frac{x_i-x_{i-1}}{2}\left(f(x_{i-1})+f(x_i)\right)

From an expression: decide whether to evaluate exactly or differentiate

If you see an integral with numeric bounds and an integrand you can antidifferentiate, FTC Part 2 gives an exact value.

If you see an integral with a variable bound (like \int_1^{g(x)} f(t)\,dt), the task is often to differentiate using FTC Part 1 and the chain rule.

Worked example: mixed skills (setup, compute, interpret)

A rate function r(t) (gallons/hour) describes net flow into a tank. You are given values at t=0,1,2,3,4 hours:

t01234
r(t)35421

The tank initially contains 10 gallons.

1) Write an expression for the net change in volume from 0 to 4:

\int_0^4 r(t)\,dt

2) Approximate using the trapezoidal rule with \Delta t=1:

\int_0^4 r(t)\,dt\approx\frac{1}{2}\left(r(0)+2r(1)+2r(2)+2r(3)+r(4)\right)

Substitute:

\frac{1}{2}\left(3+2\cdot 5+2\cdot 4+2\cdot 2+1\right)=\frac{1}{2}(26)=13

3) Approximate final volume:

V(4)\approx 10+13=23

4) Interpretation with units: 13 is in gallons because gallons/hour integrated over hours produces gallons.

Common mistakes to watch:

  • Forgetting to double the middle terms in trapezoidal rule.
  • Reporting 13 as the final volume instead of the net change.
Exam Focus
  • Typical question patterns:
    • “Given a table/graph of a rate, approximate an integral and then use it to find a quantity.”
    • “Interpret the result of an integral in context, including units.”
    • “Decide whether to use FTC evaluation, a Riemann sum approximation, or trapezoidal approximation based on the information provided.”
  • Common mistakes:
    • Using an exact FTC approach when no antiderivative is available (or when only data are given).
    • Mixing up what the integrand represents (rate vs. accumulated quantity).
    • Not matching the question’s wording: net change vs. total, approximation vs. exact.