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Exam prep · AP Calculus AB · Unit guide

AP Calculus AB Unit 8: Applications of integration

How one idea, adding up thin slices, gives averages, distance, area and volume

On the exam: Average value, motion, area between curves, volumes. 10 to 15%.

Take the AP Calculus AB quiz Make flashcards from my notes

Does the integral of velocity always tell you how far a particle traveled? It doesn't, and that mix-up costs students more points in calculus than almost anything else here. We'll cover average value, motion, area between curves and volumes, then work a full question.

The Average Value of a Function

The average value of a function on an interval is one over b minus a, times the integral from a to b of f of x. Let's try f of x equals x squared on zero to three. The integral from zero to three of x squared is twenty seven over three, which is nine. Then we multiply by one third, so the average value is three.

f_avg = 1/(b − a) ∫[a,b] f(x) dx
f(x) = x² on [0, 3]
∫[0,3] x² dx = 27/3 = 9
f_avg = (1/3)(9) = 3

Key points.

Average value is the integral divided by the interval length
Example: x² on [0, 3] has average value 3
Divide by b − a, not by 2

Displacement vs Total Distance

These two sound alike, but they answer different questions. Displacement is the integral of velocity, so backward motion cancels forward motion, and the answer can be negative. Total distance is the integral of speed, which is the absolute value of velocity, so every bit of motion counts and the answer is never negative. To compute distance by hand, you find where velocity equals zero and split the integral there.

Displacement

∫ v(t) dt
Can be positive, negative or zero
Net change in position

Total distance

∫ |v(t)| dt
Always zero or positive
Split where v(t) changes sign

The Net Change Theorem

Underneath all of this sits one principle. When you integrate a rate of change, you get the net change in the quantity. So a particle's position at time b equals its position at time a plus the integral of velocity from a to b. If it starts at x equals five and its displacement is negative three, it ends at two. The same idea works for gallons per minute or people per hour.

The integral of a rate of change over an interval gives the net change in the quantity. Net Change Theorem

Key points.

Integrating a rate gives the net change in the quantity
Final position equals starting position plus the integral of velocity
Works for any rate: water, people, temperature

Working a Motion Problem

Say velocity is t squared minus four, from t equals zero to three. The displacement is the integral of t squared minus four, which is nine minus twelve, or negative three. Now velocity is zero at t equals two, so we split. From zero to two, the distance is sixteen thirds. From two to three, it's seven thirds. Adding them gives a total distance of twenty three thirds.

v(t) = t² − 4, 0 ≤ t ≤ 3
∫[0,3] (t² − 4) dt = 9 − 12 = −3
v(t) = 0 at t = 2
∫[0,2] |v| dt = 16/3
∫[2,3] |v| dt = 7/3
distance = 16/3 + 7/3 = 23/3

Key points.

v(t) = t² − 4 on 0 ≤ t ≤ 3 gives displacement −3
Velocity is zero at t = 2, so split there
Total distance is 16/3 + 7/3 = 23/3

Area Between Two Curves

To find the area between two curves, picture a thin vertical rectangle stretching from the bottom curve up to the top curve. Its height is top minus bottom, so the area is the integral of top minus bottom with respect to x. The bounds are usually where the curves intersect. If the region is easier to slice sideways, use horizontal rectangles and integrate right minus left with respect to y.

Key points.

Area is the integral of top curve minus bottom curve
Bounds usually come from where the curves intersect
In terms of y, use right curve minus left curve

Volumes with Known Cross Sections

Some solids sit on a flat base, with slices of a known shape standing on it. You find the side length of each slice, usually top minus bottom, then plug it into the shape's area formula. A square slice has area s squared, a semicircle with diameter s has area pi over eight times s squared, and an equilateral triangle has area root three over four times s squared. Then you integrate that area.

Sketch the base region and find its bounds
Slice perpendicular to the given axis
Write the side length as top minus bottom
Plug it into the area formula A(x)
Integrate A(x) between the bounds

Discs and Washers

When you spin a region around an axis, each slice becomes a disc, so the volume is pi times the integral of R squared. If there's a gap between the region and the axis, each slice is a washer, so you use pi times the integral of R squared minus r squared. R is the outer radius and r is the inner one. Rotating about y equals negative one, you add one to each radius.

Disc: V = π ∫ R² dx
Washer: V = π ∫ (R² − r²) dx
R = outer, r = inner radius
about y = −1: R = top + 1, r = bottom + 1

Key points.

Disc: V = π ∫ R² dx when the region touches the axis
Washer: V = π ∫ (R² − r²) dx when there's a gap
Radius is distance from the axis of rotation

A Typical Free-Response Question

Here's a typical free response question. Region R sits between y equals x and y equals x squared, which meet at zero and one. For the area, integrate x minus x squared to get one sixth. Rotating about the x axis, pi times the integral of x squared minus x to the fourth gives two pi over fifteen. With square cross sections, integrating x minus x squared, squared, gives one thirtieth.

R: y = x, y = x², 0 ≤ x ≤ 1
(a) ∫[0,1] (x − x²) dx = 1/6
(b) π∫[0,1] (x² − x⁴) dx = 2π/15
(c) ∫[0,1] (x − x²)² dx = 1/30

Key points.

Region between y = x and y = x² from 0 to 1
Area 1/6, volume about the x-axis 2π/15
Square cross sections give volume 1/30

Traps to Watch For

Let's finish with the mistakes that show up again and again. For washers, you subtract the squares, so it's R squared minus r squared, not R minus r, squared. For distance, you split wherever velocity changes sign. For average value, remember to divide by the length of the interval. And for area, check which curve is on top, because curves can swap places partway through.

Key points.

Washers use R² − r², never (R − r)²
Distance needs a split where velocity changes sign
Average value divides by b − a
Check which curve is on top on each interval

Remember this

Remember that every one of these is the same move: slice the thing up, write one slice, and integrate. Now quiz yourself on a few region and motion problems to see if it sticks.

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