Unit 8 · Applications of Integration

Unit 8

Applications of Integration

~3 weeks · February into March · 5–10%

Why it's here now, and what's actually being taught

Techniques are in hand; deploy them. Conceptually this unit adds almost nothing new — the integrals are usually easy. What's being taught is a method of setup, and that method is the entire content.

The one idea

Slice. Approximate one slice as something simple. Integrate.

Every problem in this unit is: chop the object into infinitesimal pieces, write down what a single generic piece contributes, and sum. The setup is the work.

The habit that makes this unit easy

Draw and label a representative slice, every single time. Write its dimensions on the picture. That habit is worth more than any formula here — students who skip it lose track of whether a radius is x or (4 − x), and no amount of algebra recovers from that.

Area between curves

∫_a^b [top − bottom] dx or ∫_c^d [right − left] dy

Worked — with a crossing

Find the area between y = x² and y = x + 2.

x² = x + 2 → x² − x − 2 = 0 → (x−2)(x+1) = 0	find where they intersect — these are your limits
x = −1 and x = 2
test x = 0: line gives 2, parabola gives 0	check which is on top, don't assume
∫₋₁²[(x+2) − x²]dx	top minus bottom
= [x²/2 + 2x − x³/3]₋₁²
= (2 + 4 − 8/3) − (½ − 2 + ⅓) = 9/2

If the curves cross inside the interval, split there and take top−bottom separately on each piece — otherwise the regions cancel. Same issue as ∫|v| in Unit 6.

When to slice horizontally instead

Use dy when the region is more naturally described by "right curve minus left curve" — typically when a single vertical slice would change which curve bounds it partway up, forcing you to split. One dy integral often replaces two dx integrals.

The cost is that you must solve the equations for x in terms of y, which isn't always possible. Try both mentally before committing.

Check — sign and size

Area is positive. Between y = x and y = x² on [0, 1] the line is on top, so ∫₀¹(x − x²) dx = ½ − ⅓ = 1/6 ✓.

If your area comes out negative you subtracted in the wrong order. That is the single most common way to lose a point in this unit, and it is visible without redoing the integral.

Check — lumping

The gap between y = x and y = x² is widest in the middle: at x = 0.5 it is 0.5 − 0.25 = 0.25. Lump the whole region into something triangle-shaped of width 1 and height 0.25, giving about 0.125.

The exact area is 1/6 = 0.167 — the same size, which is all lumping claims. If you had computed 1.67 or 0.017, you would know immediately without rechecking a single step.

Volumes

Known cross-sections

V = ∫_a^b A(x) dx

Slice perpendicular to an axis; each slice is a slab of area A(x) and thickness dx. The cross-section might be a square, a semicircle, an equilateral triangle — whatever the problem specifies, built on a base whose width you read off the region.

Reminder — areas you'll need for cross-sections

Square on side s: s²
Semicircle on diameter s: (π/8)s² — half of π(s/2)²
Equilateral triangle on side s: (√3/4)s²
Isosceles right triangle, s as a leg: s²/2; s as the hypotenuse: s²/4

In every case s is the width of the region at that x — usually (top curve − bottom curve).

Disks and washers

The special case where the cross-sections are circles, because the solid was formed by rotating a region about a line.

Disk: V = π∫R² dx · Washer: V = π∫(R_outer² − R_inner²) dx

Rotating the region between two curves produces a solid with a hole. Each slice is a washer: outer radius R from the axis to the far curve, inner radius r from the axis to the near curve. Subtract the squares of the radii, not the radii.

Trap — the two errors that cost everyone points

One: π∫(R − r)²dx is wrong. It must be π∫(R² − r²)dx. You're subtracting areas, and (R−r)² ≠ R² − r².

Two: when the axis of rotation isn't the x-axis, the radius is the distance from the axis, not the function value. Rotating about y = 4 makes the radius (4 − f(x)); rotating about y = −1 makes it (f(x) + 1). Draw the slice and measure.

Worked — washer about a shifted axis

The region bounded by y = √x and y = x, from x = 0 to 1, is rotated about y = 2. Set up the volume.

on [0,1], √x ≥ x	check: at x = ¼, √x = ½ > ¼ ✓
the axis y = 2 is above both curves	so the farther curve gives the outer radius
y = x is farther from y = 2 → R = 2 − x	distance from the axis
y = √x is nearer → r = 2 − √x
V = π∫₀¹[(2−x)² − (2−√x)²]dx	outer squared minus inner squared

The counterintuitive step is that the lower curve gives the outer radius, because the axis is above everything. This is exactly why you draw the slice. AP problems choose shifted axes deliberately to test it.

Often taught, not on the AP

The shell method — V = 2π∫(radius)(height)dx, peeling cylindrical layers instead of stacking slices — is not in the AP Calculus course description. Many teachers cover it because it's genuinely easier for some solids (rotating about the y-axis a region defined by y = f(x), where washers would require solving for x).

Learn it if her class does; don't worry if it's skipped. Anything shells can do, washers can also do with a change of variable.

Check — known case

Revolve y = x from 0 to h and you must get a cone. Disks give V = ∫₀ʰ πx² dx = πh³/3. The cone volume you already know is ⅓πr²h, and here r = h — the same thing ✓.

Any volume setup that cannot reproduce a cone is being applied wrongly, and testing it on a cone is faster than re-deriving the setup.

Arc length

L = ∫_a^b √(1 + (dy/dx)²) dx

Why — it's Pythagoras on an infinitesimal triangle

Take a tiny piece of curve. It's essentially straight, so it's the hypotenuse of a right triangle with legs dx and dy:

ds² = dx² + dy² → ds = √(dx² + dy²) = √(1 + (dy/dx)²) dx

Factoring dx out of the root is the only step. Then sum the ds pieces.

Worth knowing: ds² = dx² + dy² is the metric, and generalizing it to curved spaces is literally how differential geometry — and then general relativity — begins. The arc length formula is a first glimpse of a very deep object.

Trap — arc length integrals are usually not elementary

√(1 + (f′)²) rarely has a nice antiderivative. Most arc length problems are set-up-only, or expect a calculator. Don't let her burn ten minutes trying to integrate it by hand. The AP asks for the setup far more often than the value.

Accumulation in context

The most common free-response format in the whole exam: a rate is given, in words or as a graph, and the questions are all Unit 6's net-change theorem wearing different clothes.

Asked	Answer
How much accumulated between a and b?	∫_a^b rate dt
How much is there at time b?	initial + ∫_a^b rate dt
Average rate over [a,b]	[1/(b−a)]∫_a^b rate dt
When is the amount greatest?	where the net rate changes from + to −
Two rates, in and out	∫(in − out) dt
Work with a varying force	∫F(x)dx

Trap — units and the initial condition

Two points get lost here repeatedly. Always state units — the integral of gallons-per-hour over hours is gallons, and saying so earns a point. And the integral gives change, never total: if the question asks how much is in the tank, add the starting amount.

Where this leads — probability

Next stop — continuous probability. A density is a function; probability is its integral; expected value is ∫x·p(x)dx; variance is another integral. The bell curve is e^−x²/2 normalized, and its non-elementary antiderivative is exactly why every statistics text has a Z-table.

Then — stochastic calculus, which does calculus on paths that are continuous everywhere and differentiable nowhere. Those paths are Weierstrass's function from Unit 1, made physical. Itô's lemma is the chain rule corrected for the fact that the path has no derivative, and the Black–Scholes equation is a PDE built on it. Unit 1's "lamentable plague" turns out to be the foundation of modern finance.

Beyond BC · what a college course does here

Three standard applications the AP framework omits. All three are the same slice-and-sum method, so none of them is new work.

1 · Surface area of revolution. Rotating a curve sweeps out a surface; each arc-length element ds traces a thin band of circumference 2πy:

S = ∫ 2πy ds = ∫ 2πy √(1 + (dy/dx)²) dx

The ds — not dx — is the whole point, and it's why arc length had to come first. Use this to finish Gabriel's Horn honestly rather than by inequality.

2 · Work. Force times distance, when the force varies: W = ∫F(x)dx. Two classics: a spring obeying Hooke's law F = kx gives W = ½kx²; and pumping water out of a tank means integrating (weight of a slice) × (distance that slice must travel), which is where students learn to be careful about which distance.

3 · Centroids and Pappus. The centroid of a region is its balance point, x̄ = (1/A)∫x·(height)dx. And then a genuinely surprising theorem:

Pappus: V = 2π·r̄·A

The volume of a solid of revolution equals the area of the region times the distance its centroid travels. A torus made by revolving a circle of radius r about an axis R away has volume 2πR·πr² — no integration at all. Pappus stated this around 300 AD, thirteen centuries before calculus existed.

Beyond BC · the Russian approach — one region, four methods

Rather than a hundred volume problems, the tradition would take one region and attack it every possible way, because agreement between methods is a proof, and disagreement locates your error.

Take the region under y = x² from 0 to 1, rotated about the y-axis. Find the volume four ways.

Washers in y: solve x = √y, slice horizontally. V = π∫₀¹[1² − (√y)²]dy = π∫(1 − y)dy = π/2.
Shells in x: V = 2π∫₀¹x·(1 − x²)dx = 2π[x²/2 − x⁴/4] = π/2. ✓
Cylinder minus solid: the full cylinder is π(1)²(1) = π; subtract the solid under the curve, π∫y dy = π/2. Leaves π/2. ✓
Pappus: compute the region's area and centroid, multiply. ✓

Four routes, one answer. The habit being built is not volume computation — it's the expectation that a correct result should be reachable more than one way, and the reflex to check rather than to hope. That reflex is worth more on an exam than any formula on this page.

Formula sheet

Unit 8 — Applications of Integration

The method

Slice → approximate one slice → integrate. Draw and label a representative slice every time.

Area

∫(top − bottom)dx or ∫(right − left)dy
Find intersections first — they're the limits. Split wherever the curves cross.

Volume

Cross-sections: V = ∫A(x)dx
Disk: π∫R²dx · Washer: π∫(R² − r²)dx — never (R−r)²
Radius = distance from the axis. Shifted axis y = k ⟹ radius |k − f(x)|.
Square s² · semicircle on diameter s (π/8)s² · equilateral triangle (√3/4)s²

Arc length (BC)

L = ∫√(1 + (dy/dx)²)dx — from ds² = dx² + dy²
Usually not elementary. Expect setup-only or calculator.

Accumulation

amount at b = initial + ∫_a^brate dt
Net rate in − out; maximum where the net rate crosses zero downward
State units. Add the initial condition.

Practise Unit 8 → Fresh problems from this unit, with worked solutions.