Applications of Integration: AP Calc AB Unit 8

Why this unit matters

These applications show how integration answers total-amount questions that derivatives alone cannot.

What you will learn

Compute net and total change from rate functions on intervals.
Find area between curves using intersection bounds.
Model accumulation in position, revenue, and other applied settings.

Understand the core ideas

Applications of integration ask for totals, not just local rates. If a rate r(t) is known, total change from t = a to t = b is $\int_{a}^{b} r(t) \, dt$ . In AP Calculus AB, this idea appears in displacement from velocity, total revenue from marginal revenue, and accumulated amount from inflow or outflow rates. You should always interpret whether the integral gives net change or total amount, because sign can change meaning.

Area between curves is another core AB application. If one function is above another on an interval, area is integral of top minus bottom. Correct bounds often come from intersection points, so solve equations first. For average value on [a, b], use (1/(b-a)) times $\int_{a}^{b} f(x) \, dx$ . This gives a representative height whose rectangle has the same area as the region under the curve. In context problems, finish by stating units and practical interpretation, not only the computed number.

Key terms

Net change: The overall signed change in a quantity over an interval, computed by integrating its rate.
Area between curves: The positive geometric area found by integrating upper function minus lower function on the interval.
Average value of a function: The quantity (1/(b-a)) * $\int_{a}^{b} f(x) \, dx$ , representing mean height on an interval.
Displacement: Net change in position, found by integrating velocity over time.

Find volume with square cross sections

A solid has base region under y = x from x = 0 to x = 2. Cross sections perpendicular to the x-axis are squares whose side length is y.

At position x, the square side is y = x, so its cross-sectional area is $A(x) = x^2$ .
Set up the volume integral: V = $\int_{0}^{2} x^2 \, dx$ .
Use the antiderivative $x^3/3$ .
Evaluate: $V = 2^3/3 - 0^3/3 = 8/3$ .

Result: The solid has volume 8/3 cubic units.

A common misconception

Claim: A volume problem always integrates a length directly.

Correction: Volume by cross sections integrates area. First write A(x) from the cross-section shape, then compute V = integral of A(x) over the base interval.

Lessons in this unit

Accumulation from ratesConvert a known rate function into total change over time.
Area between two curvesSet up top-minus-bottom integrals with correct bounds.
Volume from cross sectionsIntegrate known cross-sectional area formulas across an interval.
Average value of a functionUse integral mean value to summarize behavior on an interval.

Study task

Two functions model cost and revenue rates. Integrate each over the same interval, compare totals, and interpret the difference in context.

Unit checkpoint

v(t) = 6t - t^2

on 0 <= t <= 4, what is the displacement on [0, 4]?

Displacement is $\int_{0}^{4} (6t - t^2) \, dt$ $= [3t^2 - t^3/3]$ from 0 to 4 = 48 - 64/3 = 80/3.

Applications of Integration