AP Calculus BC - Unit 8 Study Guide (Edited)
AP Calculus BC • Unit 8

Applications of Integration

Exam Weight: 10–15%
8.1

Average Value of a Function

How do you find the average height of a curve over an interval? You "smush" the area into a rectangle.

favg a b
The Formula
favg = [ 1 / (b - a) ] · ∫ab f(x) dx

"Integral divided by Interval Width"

AP Exam Moves
  • Always show (b-a) as the width.
  • Units: average value has the same units as f(x).
  • Common FRQ follow-up: find c where f(c)=favg.
Example

Find the average value of f(x) = x² on [0, 3].

  1. Interval Width: b - a = 3 - 0 = 3.
  2. Integrate: ∫03 x² dx = [x³ / 3]03 = 27/3 - 0 = 9.
  3. Divide: Average = 9 / 3 = 3.
AP-Style Follow-Up (Mean Value for Integrals)

Find a value c in [0,3] such that f(c)=favg.

  1. Set x² = 3.
  2. Solve: x = √3 (the only solution in the interval).

On FRQs: one equation + one value in the interval = full credit.

Common Mistake
Forgetting to divide by (b-a). If you only compute ∫ f(x)dx, that’s the area, not the average height.
Mini Practice

1) Find favg for f(x)=\sin x on [0, \pi].

2) For the same function, find c where f(c)=favg (give one value).

Show Answers

1) favg = \frac{1}{\pi-0}\int_0^\pi \sin x\,dx = \frac{1}{\pi}[ -\cos x ]_0^\pi = \frac{1}{\pi}(2)=\frac{2}{\pi}.

2) Solve \sin c = \frac{2}{\pi}. One solution is c=\arcsin\left(\frac{2}{\pi}\right) (in [0,\pi]).

8.2

Position, Velocity, Acceleration (Integrals)

In Unit 4, we differentiated position to get velocity. Now, we integrate velocity to get position.

Net vs. Total
Concept Formula Meaning
Displacement ∫ab v(t) dt Net change in position. (Can be 0 if you return home).
Total Distance ∫ab |v(t)| dt Total ground covered. (Odometer reading). Always positive.
Current Position s(t) = s(a) + ∫at v(x) dx Start Position + Displacement.
AP Exam Moves
  • Find where v(t)=0 (turning points) before doing total distance.
  • Sign matters: negative velocity means moving backward (displacement can be negative).
  • Units: if v is m/s and t is seconds, the integral is meters.
Example: Displacement vs Total Distance (Velocity Changes Sign)

Let v(t)=t²-4 on 0 ≤ t ≤ 3.

  1. Where does it change sign? t²-4=0 → t=2.
  2. Displacement: ∫03(t²-4)dt = [t³/3 - 4t]03 = (9 - 12) = -3.
  3. Total Distance: split at t=2:
    ∫02(4-t²)dt + ∫23(t²-4)dt

    Because t²-4 is negative on [0,2] and positive on [2,3].

Common Mistake
Doing ∫|v| without splitting at the zeros of v. Absolute value is not “magic”—you must handle sign changes.
Mini Practice

1) If s(0)=5 and v(t)=2t, find s(3).

2) Write (don’t compute) an expression for total distance traveled if v(t)=\sin t on [0, \pi].

Show Answers

1) s(3)=5+∫032t\,dt = 5+[t²]03=14.

2) ∫0\pi|\sin t|\,dt (here it’s all positive, so =∫_0^\pi \sin t\,dt).

8.4 - 8.6

Area Between Curves

To find the area between two functions, we slice it into rectangles. Height = Top - Bottom.

Top Curve Bottom Curve dx

Vertical Slices (dx)

Top Function minus Bottom Function

Area = ∫ab [ Top(x) - Bot(x) ] dx

Horizontal Slices (dy)

Right Function minus Left Function

Area = ∫cd [ Right(y) - Left(y) ] dy
AP Exam Moves
  • Step 0: find intersection points (solve f(x)=g(x)) to get limits.
  • Pick the easier variable: if one way needs splitting into 2+ integrals, try the other.
  • Area is positive: use Top-Bottom (or Right-Left) so you don’t get a negative answer.
Example (dx): Area Between y=x and y=x²
  1. Intersections: x = x² → x(x-1)=0 → x=0,1.
  2. Top vs Bottom on [0,1]: x is above x².
  3. Area: ∫01(x - x²)dx.

AP loves “set up the integral” questions. This is already full credit for setup.

Example (dy): When horizontal slices are cleaner

Region bounded by x=y² and x=2y.

  1. Intersections: y² = 2y → y(y-2)=0 → y=0,2.
  2. Right - Left: right is x=2y, left is x=y².
  3. Area: ∫02(2y - y²)dy.
Common Mistake
Picking dy but forgetting to rewrite curves as x = ... functions of y. For dy, you need right/left in terms of y.
Mini Practice

1) Set up (don’t compute) the area between y=\ln x and y=1 on [1,e].

2) If you switch to dy, what would you need first?

Show Answers

1) On [1,e], top is 1 and bottom is \ln x, so Area = ∫1e(1-\ln x)\,dx.

2) You’d need x as functions of y (solve y=\ln x → x=e^y) and determine right/left limits in y.

8.7 - 8.8

Volumes with Cross Sections

Imagine a 3D solid built on a base region R. Volume is the integral of area of each slice.

Volume = ∫ab Area(x) dx
Fast Setup Recipe
  1. Sketch the base region and label Top(x) and Bottom(x).
  2. Write the slice width: s = Top(x) - Bottom(x).
  3. Convert s into an area formula (square, triangle, semicircle, etc.).
  4. Integrate Area(x) over the correct limits.
Common Area Formulas (Memorize)

If the base of the slice is s = Top(x) - Bot(x):

  • Square: A = s²
  • Semicircle: diameter = s → A = (π / 8)s²
  • Equilateral Triangle: side = s → A = (√3 / 4)s²
  • Isosceles Right Triangle: legs = s → A = (1/2)s²
  • Rectangle: A = s · h (height usually given).
Example: Squares on the region between curves

Base region is between y=x (top) and y=0 (bottom) from x=0 to x=2. Cross sections perpendicular to the x-axis are squares.

  1. Slice width: s = Top - Bottom = x - 0 = x.
  2. Area of square: A(x)=s²=x².
  3. Volume: V=∫02x²\,dx.

On FRQs, labeling s clearly is often the difference between 3/4 and 4/4.

Common Mistake
Using Top+Bottom instead of Top-Bottom. The slice width s is a distance, so it should be positive.
Mini Practice

1) Base is between y=4-x² and y=0. Cross sections are semicircles with diameter in the base. Set up V.

Show Answer

Intersections: 4-x²=0 → x=±2. Diameter s= (4-x²)-0. Semicircle area: A=(π/8)s².

V = ∫-22 (π/8)(4-x²)² \, dx.

8.9 - 8.10

Rotational Volume: The Disc Method

Use disc method when the region is flush against the axis of rotation (no hole).

Axis of Rotation R(x)
V = π ∫ab [ R(x) ]² dx

Think: sum of circle areas (πR²).

AP Exam Moves
  • Radius is a distance to the axis. If rotating around y=k, radius becomes |f(x)-k|.
  • Write: “Radius = (top curve) − (axis)” (or “axis − curve”) so graders see the geometry.
Example: Rotate y=x about the x-axis (disc)

Region under y=x from x=0 to x=2 rotated about the x-axis.

  1. Radius: R(x)=x.
  2. Volume: V=π∫02x²dx.
Common Mistake
Squaring the integral (like [∫R(x)dx]²). You must integrate πR(x)², not square at the end.
Mini Practice

1) Set up the volume when the region under y=\sqrt{x} from 0 to 4 rotates about the x-axis.

Show Answer

R(x)=\sqrt{x} so V=π∫04(\sqrt{x})²dx = π∫04x\,dx.

8.11 - 8.12

Rotational Volume: The Washer Method

Use washer method when there is a gap between the region and the axis. The slice looks like a donut.

Axis of Rotation Big R Little r
V = π ∫ab ( [R(x)]² - [r(x)]² ) dx

"Big Radius Squared minus Little Radius Squared"

Big Radius (R)

Distance from Outer Curve to Axis.

Little Radius (r)

Distance from Inner Curve to Axis.

Warning: Calculate R² - r². Do NOT calculate (R - r)². That is Algebra murder.

AP Exam Moves
  • Write radii as distances: R(x)=|top-axis|, r(x)=|bottom-axis| (depending on picture).
  • If rotating about y=k, your radii become |f(x)-k| and |g(x)-k|.
Example: Rotate the region between y=2 and y=x about the x-axis

Region between y=2 (top) and y=x (bottom) from x=0 to x=2, rotated about the x-axis.

  1. Outer radius: R(x)=2.
  2. Inner radius: r(x)=x.
  3. Volume: V=π∫02(2² - x²)\,dx.
Common Mistake
Mixing up which curve is “outer.” The outer radius is the one farther from the axis at that x-value.
Mini Practice

1) Set up the volume when the region between y=x² and y=4 rotates about the x-axis.

Show Answer

Intersections: x²=4 → x=±2. Outer radius R=4, inner radius r=x².

V=π∫-22(4²-(x²)²)\,dx = π∫-22(16-x⁴)\,dx.

8.13

Arc Length BC Only

Arc length is the physical length of a curve. We approximate with tiny hypotenuses and integrate.

The Arc Length Formula
L = ∫ab √[ 1 + (f'(x))² ] dx

You MUST take the derivative first, then plug into the formula.

AP Exam Moves
  • Most arc length integrals are ugly. On AP, you may only need the setup, or you may be told to approximate.
  • Perfect-square check: if 1+(f')² becomes a perfect square, the integral gets much easier.
  • Calculator note: If asked to “approximate,” use a calculator and show the integral you evaluated.
Example (Setup Only): Arc length of y = x³ on [0,1]
  1. Derivative: f'(x)=3x².
  2. Arc length: L=∫01√(1+(3x²)²)\,dx = ∫01√(1+9x⁴)\,dx.

This is exactly the kind of FRQ where they give full credit for correct setup and may ask for a numerical approximation.

Common Mistake
Plugging f(x) into the formula instead of f'(x). Arc length uses the derivative.
Mini Practice

1) Set up (don’t compute) the arc length of y=\ln x on [1,e].

Show Answer

f'(x)=1/x, so L=∫1e√(1+(1/x)²)\,dx.

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