AP Calculus AB Unit 8: Applications of Integration Notes

Introduction

An applied integral begins with a small contribution: rate times time, height times width, or cross-sectional area times thickness.

The diagram decides the slicing variable, bounds, subtraction order, and radius expressions before any antiderivative is computed.

A complete setup names the accumulated quantity and carries its units from the integrand to the final interpretation.

Current exam guidance: May 2027 format: 42 multiple-choice questions and 6 free-response questions. Calculator availability changes by part; the reader marks calculator decisions separately from mathematical justification.

8.1

Finding the Average Value of a Function on an Interval

Why does average value divide accumulated output by interval length?

Short answer

It is the constant height whose rectangle over the interval has the same signed accumulation as the function.

Know firstDefinite integrals · Interval length · Units

Replace a varying output with an equivalent constant height

The integral totals infinitely many thin output-by-input contributions. Dividing that total by b−a spreads it evenly across the interval. For f(x)=x²+1 on [0,2], the total is 14/3 and the average height is 7/3.

This is an input-weighted average, not the output at the midpoint. If f is continuous, the Mean Value Theorem for Integrals guarantees at least one input where the function reaches the average height.

Formula

Formula and conditions

f_{\mathrm{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx

A constant output with the same total accumulation as f.

Use when: a<b and f is integrable; continuity is needed only to guarantee f(c)=f_avg.
Do not assume: Do not divide by the number of points visible in a table.

One constant height matches the total accumulation

The horizontal line y=7/3 has the same area over [0,2] as f(x)=x²+1, so it represents the function’s average height.

The interactive version of this figure is part of the full study guide.

One constant height matches the total accumulation: values

Case	Result
Total accumulation	$14/3$
Interval length	2
Average value	$7/3$

Build the integral from one slice or rate

Compute the interval length b−a.
Integrate over the same bounds.
Divide and attach the function’s output units.
Distinguish the average height from an input c.

Keep these three ideas

Integral gives total.
Interval length converts total to average.
Continuity guarantees the height occurs.

Keep reading Unit 8: Applications of Integration

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8.1 Finding the Average Value of a Function on an IntervalRead

8.2 Connecting Position, Velocity, and Acceleration Using Integrals

8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

8.4 Finding the Area Between Curves Expressed as Functions of x

8.5 Finding the Area Between Curves Expressed as Functions of y

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

8.7 Volumes with Cross Sections: Squares and Rectangles

8.8 Volumes with Cross Sections: Triangles and Semicircles

8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

8.10 Volume with Disc Method: Revolving Around Other Axes

8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

8.12 Volume with Washer Method: Revolving Around Other Axes

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