Applications of Integration

What this unit covers

The topics below follow the published Calculus BC course framework for Unit 8. This unit is worth 5–10% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Average Value of a Function12 min · 3 objectivesCompute the average value of a function over an interval · Distinguish the average value of f from the average rate of change · Relate average value to the Mean Value Theorem for Integrals
• Displacement, Total Distance and Net Change15 min · 3 objectivesDistinguish displacement from total distance and choose the right integral for each · Apply the Net Change Theorem to recover an amount from a rate and an initial value · Interpret an integral of a rate in context, with units
• Area Between Curves14 min · 3 objectivesSet up an integral for the area between two curves · Determine which function is on top and find the bounds from intersections · Choose between integrating with respect to x or y
• Volumes of Solids15 min · 3 objectivesCompute volumes by the disk and washer methods · Set up volumes of solids with known cross sections · Select the correct method from the geometry of the region
• Arc Length13 min · 3 objectivesCompute the length of a curve using the arc-length integral · Set up arc length for a function y = f(x) · Recognize arc length as an accumulation of tiny hypotenuses

Formulas in Unit 8

Every term in Unit 8

All 24 terms we publish for Applications of Integration, 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.

Arc length of y = f(x)

L = ∫√(1 + (dy/dx)²)dx over the interval. The integrand is almost never elementary, so these are usually calculator questions.

Arc length of a parametric curve

L = ∫√((dx/dt)² + (dy/dt)²)dt, which reduces to the Cartesian form when the parameter is x.

Volume by known cross sections

V = ∫A(x)dx with A(x) the cross-sectional area. Include π only if the cross sections are circular.

Average value of a function

f_avg = (1/(b − a))∫ₐᵇ f(x)dx — the height of the rectangle with the same area over the interval.

Total distance from a velocity function

∫|v(t)|dt, which requires splitting at every t where v changes sign. Displacement is ∫v(t)dt without absolute value.

Area between two curves

∫(top − bottom)dx, splitting at each intersection where the curves swap position.

Volume by disks and washers

π∫r²dx for disks, π∫(R² − r²)dx for washers. Subtract the squares, never square the difference.

Choosing dx or dy

Integrate perpendicular to the slices — usually dx about a horizontal axis and dy about a vertical one.

Cross-section area formulas

Square s², equilateral triangle (√3/4)s², semicircle (π/8)s² with s the diameter. Check whether s is a side or a diameter.

Displacement vs total distance

Displacement is ∫v dt; total distance is ∫|v|dt, requiring a split wherever v changes sign.

Arc length comes from Pythagoras

A piece of curve has run dx and rise dy, so its length is √(1 + (dy/dx)²)dx. The parametric form is the same idea in t.

Arc-length integrands rarely antidifferentiate

On a calculator section the expected answer is a setup plus a number; on a non-calculator section, the setup alone.

Write the integral before evaluating

Rubrics award the setup separately from the value, so an unexplained decimal earns a fraction of the credit.

Vertical slices mean top minus bottom

Integrating dx runs each slice from the lower curve to the upper one, both as functions of x.

Horizontal slices mean right minus left

Integrating dy runs each slice from the left curve to the right one, both as functions of y.

Switch variables before splitting

Two integrals doubles the chance of an algebra error; switching usually costs only solving for the inverse.

Test which curve is on top

Substitute a point inside the interval. A negative area means the order was reversed.

Upper curve can switch

With more than two intersections the region must be split at the crossing, and each piece keeps the upper curve on top.

Net change from a rate in an applied context

∫ₐᵇ f′ = f(b) − f(a). The integral of a rate is a change in the quantity.

Amount equals initial plus change

Reporting the integral alone as the amount is the most common applied-integration error, because the initial value is in a different sentence.

In minus out

Net rate R − S; the amount is greatest where that changes from positive to negative, which still requires comparing with the endpoints.

Average value divides by the width

(1/(b − a))∫ₐᵇ f. Omitting the division reports an accumulated total as an average.

Average value keeps the function's units

The average of a velocity is a velocity, not a distance.

MVT for integrals

A continuous function attains its average value somewhere on the interval — the integral analogue of the Mean Value Theorem.

What examiners penalize here

• Read the wording precisely: "average value of the velocity" means integrate v and divide by the time interval, while "average acceleration" means [v(b) − v(a)]/(b − a). One is an integral average, the other an endpoint average — pick the right formula.
• Read the verb. "How far did it travel" is total distance; "what is its position" and "how far from its starting point" are displacement questions. The integrals differ by absolute-value bars and the answers differ by a lot.
• When the region is easier to slice horizontally — for instance bounded by curves given as x = h(y) — integrate with respect to y using right minus left. Choosing the orientation that avoids splitting the region saves time on free response.
• For cross-section problems, first decide whether the region’s height gives the *side* or the *diameter* of the shape. Semicircle and some square-on-the-diagonal setups use the height as a diameter, which changes the area formula — misreading this is a common lost point.
• Most arc-length integrals cannot be evaluated by hand, so on the calculator section, set up ∫√(1 + (dy/dx)²) dx precisely and let the calculator finish. The setup — correct derivative, correct bounds — is what the rubric rewards.

Practice Calculus BC

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 10 units of AP Calculus BC

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Fundamental Properties
3. Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
4. Unit 4 · Contextual Applications of Differentiation
5. Unit 5 · Analytical Applications of Differentiation
6. Unit 6 · Integration & Accumulation of Change
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration
9. Unit 9 · Parametric, Polar & Vector-Valued Functions
10. Unit 10 · Infinite Sequences & Series