Integration & Accumulation

What this unit covers

The topics below follow the published Calculus AB course framework for Unit 6. This unit is worth 15–20% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Riemann Sums & the Definite Integral14 min · 3 objectivesApproximate area under a curve with left, right, and midpoint Riemann sums · Interpret the definite integral as a limit of Riemann sums · Predict whether a sum overestimates or underestimates based on monotonicity
• Antiderivatives & the Fundamental Theorem15 min · 3 objectivesFind antiderivatives using the reverse power rule and basic formulas · Evaluate definite integrals with the Fundamental Theorem of Calculus · Include the constant of integration for indefinite integrals
• U-Substitution14 min · 3 objectivesRecognize an integrand as f(g(x))·g′(x), the reverse of the chain rule · Carry out u-substitution for indefinite and definite integrals · Change the limits of integration when substituting in a definite integral
• Accumulation Functions14 min · 3 objectivesInterpret an integral with a variable upper limit as an accumulation function · Differentiate an accumulation function with the Fundamental Theorem (Part 1) · Use accumulation to find a quantity from its rate of change
• Riemann Sum Variants, and Which Way They Err15 min · 3 objectivesCompute left, right, midpoint, and trapezoidal approximations · Determine whether an approximation over- or under-estimates from monotonicity or concavity · Handle unequal subintervals, as tables usually provide
• The Two Fundamental Theorems, Kept Straight15 min · 3 objectivesState both parts of the Fundamental Theorem of Calculus and their distinct uses · Differentiate an accumulation function, including with a variable inside the upper limit · Recognize which part a question requires
• Definite Integrals from Graphs and Properties14 min · 3 objectivesEvaluate a definite integral by computing signed geometric area · Apply the additivity, reversal, and constant-multiple properties · Combine given integral values to obtain a requested one

Formulas in Unit 6

Every term in Unit 6

All 47 terms we publish for Integration & Accumulation, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Riemann sums

Left, right, midpoint and trapezoidal approximations of area. Left sums underestimate an increasing function; right sums overestimate it.

u-substitution

Choose u so that du appears (up to a constant) in the integrand. For a definite integral, either change the limits or convert back before evaluating.

Trapezoidal rule

(Δx/2)[f(x₀) + 2f(x₁) + … + 2f(xₙ₋₁) + f(xₙ)]. Overestimates where the function is concave up.

Definite integral as a limit of Riemann sums

∫f dx is the limit as the subinterval width goes to zero, which is why it represents exact accumulated change.

Fundamental Theorem of Calculus part 1

d/dx ∫ₐˣ f(t)dt = f(x). With a variable upper limit g(x), the chain rule gives f(g(x))·g′(x).

Fundamental Theorem of Calculus part 2

∫ₐᵇ f(x)dx = F(b) − F(a) for any antiderivative F. Connects accumulation to antidifferentiation.

Accumulation function

g(x) = ∫ₐˣ f(t)dt increases where f is positive and has a maximum where f changes from positive to negative.

Properties of definite integrals

Reversing limits negates the integral, an integral from a to a is zero, and integrals split at any interior point.

Common antiderivatives

∫xⁿdx = xⁿ⁺¹/(n+1) for n ≠ −1, ∫(1/x)dx = ln|x| + C, ∫eˣdx = eˣ + C. The absolute value in the log matters.

Net change theorem

∫ₐᵇ f′(x)dx = f(b) − f(a). The integral of a rate gives the total change in the quantity.

Interpreting an integral in context

"The total volume of water that entered the tank between t = 0 and t = 5 minutes was 40 liters." Include what accumulated, over what interval, with units.

Choosing u in a substitution

Pick the inner function whose derivative already appears, up to a constant multiple. If nothing cancels, the substitution is wrong.

Definite integral with substitution

Either convert the limits to u values or convert back to x before evaluating. Mixing limits and variables is a guaranteed error.

Splitting an integral at a sign change

For total distance or area between curves, split wherever the integrand changes sign, then add the absolute values.

Riemann sum from a table

The subintervals may have unequal widths. Multiply each function value by its own width rather than assuming a uniform Δx.

Over- or underestimate justification

Say why: a left sum underestimates because the function is increasing; the trapezoid rule overestimates because the graph is concave up.

Accumulation function analysis

g(x) = ∫ₐˣ f(t)dt has g′ = f, so g increases where f is positive and is concave up where f is increasing.

Left sum with an increasing function

Underestimates, because the left endpoint is the smallest value on each subinterval. Monotonicity, not concavity, decides this.

Right sum with an increasing function

Overestimates. Reverse both conclusions for a decreasing function.

Trapezoidal rule and concavity

Concave up makes each chord lie above the curve, so the trapezoidal sum overestimates. Concave down reverses it.

Midpoint rule and concavity

Runs opposite to the trapezoid: on a concave-up function the midpoint rule underestimates.

Which property decides over or under

Monotonicity for left and right sums; concavity for trapezoidal and midpoint. Naming the property is part of the justification.

Unequal subintervals are normal

Table data rarely has equal spacing. Compute each width separately; the (h/2) shortcut is valid only for equal widths.

Trapezoid on one subinterval

Average the two heights and multiply by THAT subinterval’s width: [(f(x₀) + f(x₁))/2](x₁ − x₀).

Midpoint sums need midpoints

A midpoint approximation from a table is only possible when the table happens to contain the midpoint values.

FTC Part 1 is for evaluating

∫ₐᵇ f = F(b) − F(a). Turns an integral into arithmetic on an antiderivative.

FTC Part 2 is for differentiating

If g(x) = ∫ₐˣ f(t)dt then g′(x) = f(x). No antiderivative is ever computed.

FTC Part 2 with a chain rule

d/dx ∫ₐ^(u(x)) f(t)dt = f(u(x))·u′(x). Substitute the upper limit, then multiply by its derivative.

Variable in the lower limit

The sign flips: d/dx ∫ₓ^b f(t)dt = −f(x), because swapping limits negates an integral.

Do not integrate when asked to differentiate

If a question shows an integral and asks for a derivative, use Part 2. The exam includes integrands with no elementary antiderivative to enforce this.

g′ = f for an accumulation function

So g increases where f is positive, and g has a local extremum where f changes sign. Shift every Unit 5 question up one derivative.

g″ = f′ for an accumulation function

So g is concave up where f is increasing, and g has an inflection point where f changes direction.

g(a) = 0 always

The integral from a to a is zero, and this is frequently the only value of g you are given.

Definite integral as signed area

Area above the axis counts positive, below counts negative. For a graph of line segments and semicircles this is faster than antidifferentiation.

Semicircle area contribution

A semicircle of radius r contributes (1/2)πr². Check the sign against which side of the axis it lies on.

Additivity of integrals

∫ₐᵇ + ∫ᵇᶜ = ∫ₐᶜ, valid for any ordering of a, b and c.

Reversal property

∫ᵇᵃ f = −∫ₐᵇ f. The property most often forgotten, and the one recombination questions are built around.

Integral of a constant

∫ₐᵇ c dx = c(b − a). The most-dropped term in recombination problems.

No product rule for integrals

∫fg dx is not ∫f·∫g. If a question seems to need that, the intended method is substitution.

Recombining given integral values

Write the target as a signed combination of what you are given using additivity and reversal. Work symbolically before substituting.

u-substitution changes the limits

When substituting in a definite integral, convert the limits to u values — or convert back to x before evaluating. Mixing them is a silent error.

Choosing u

Pick the inner function whose derivative is present, up to a constant factor. If nothing matches, substitution is not the method.

Antiderivative of 1/x

ln|x| + C, and the absolute value matters for intervals where x is negative.

Splitting at a sign change

For ∫|f| or for total distance, split at every zero of f and add magnitudes rather than integrating across.

Riemann sum from a graph

Read heights off the curve at the required endpoints or midpoints, then multiply by widths. Do not attempt to find a formula first.

Definite integral is a number

It has no x in it, and its value does not depend on the variable of integration. An answer containing x signals a missed evaluation.

Net area versus total area

The integral gives net signed area; total area requires the integral of the absolute value. Questions distinguish these deliberately.

What examiners penalize here

• AP problems often give data in a table and ask for a specific Riemann sum. Read carefully whether left, right, or midpoint endpoints are required, and note that subintervals may have *unequal* widths — compute each rectangle’s width separately then.
• Verify any antiderivative by differentiating it back — if F′ returns the integrand, F is correct. This quick check catches reverse-power-rule slips (like forgetting to divide by the new exponent) before they cost points.
• On the exam, converting the limits to u-values is usually faster and less error-prone than back-substituting. Write the new limits explicitly next to the integral so you evaluate the u-antiderivative at the correct endpoints.
• For "how much is there at time b" problems, use final value = initial value + ∫[a to b] rate dt. Integrating the rate gives only the *change*; you must add the starting amount to get the total. This structure appears on nearly every AP integral free-response.
• If a question shows an integral and asks for a **derivative**, reach for Part 2 and do not integrate. Students who evaluate the integral first can still reach the right answer on simple integrands, but they lose the time and they fail entirely when the integrand has no elementary antiderivative — which the exam includes deliberately.

Practice Calculus AB

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 8 units of AP Calculus AB

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Rules
3. Unit 3 · Composite & Implicit Differentiation
4. Unit 4 · Contextual Applications
5. Unit 5 · Analytical Applications
6. Unit 6 · Integration & Accumulation
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration