What you'll learn
This revision guide covers the testable applications of integration in AP Calculus AB. You will learn to use definite integrals to calculate areas between curves, volumes of solids of revolution, and solve accumulation problems involving rates of change. These applications form the foundation of integral calculus and appear regularly on the AP exam.
Key terms and definitions
Definite integral — the signed area between a curve and the x-axis over an interval [a, b], evaluated as F(b) - F(a) where F is an antiderivative
Accumulation function — a function defined by a definite integral with a variable upper limit, representing total change from a reference point
Solid of revolution — a three-dimensional solid created by rotating a region in the plane about a horizontal or vertical axis
Disk method — a technique for finding volumes by slicing a solid perpendicular to the axis of revolution, creating circular cross-sections
Washer method — an extension of the disk method used when the solid has a hollow center, requiring subtraction of inner radius from outer radius
Cross-sectional area — the area of a two-dimensional slice through a three-dimensional solid, perpendicular to a given axis
Net change — the total change in a quantity over an interval, calculated by integrating its rate of change
Average value of a function — the constant height of a rectangle with the same area as the region under a curve over an interval, given by (1/(b-a))∫[a to b]f(x)dx
Core concepts
Area between curves
When finding the area between two curves, identify which function is greater on the interval of integration.
For curves y = f(x) and y = g(x) where f(x) ≥ g(x) on [a, b]:
Area = ∫a to b - g(x))dx
Key steps:
- Sketch both curves to visualize the region
- Find intersection points by solving f(x) = g(x)
- Determine which function is on top
- Set up the integral with (upper - lower)
- Evaluate the definite integral
If the curves intersect within the region, split the integral at intersection points and adjust which function is subtracted in each interval.
For horizontal slicing (integrating with respect to y):
Area = ∫c to d - f(y))dy
where g(y) is the rightmost function and f(y) is the leftmost function on the interval [c, d].
Volumes of solids with known cross-sections
When a solid has known cross-sectional shapes perpendicular to an axis, the volume is found by integrating the area of the cross-section.
General formula:
V = ∫[a to b]A(x)dx
where A(x) is the area of the cross-section at position x.
Common cross-sectional shapes:
- Square: A(x) = [s(x)]², where s(x) is the side length
- Rectangle: A(x) = length × width, both expressed as functions of x
- Semicircle: A(x) = (π/2)[r(x)]², where r(x) is the radius
- Equilateral triangle: A(x) = (√3/4)[s(x)]², where s(x) is the side length
- Isosceles right triangle: A(x) = (1/2)[s(x)]², where s(x) is the leg length
The base of the solid typically lies in a region bounded by curves. Express the dimensions of each cross-section in terms of the distance between these curves.
Volumes of revolution — disk method
When rotating a region about a horizontal or vertical axis with no gap between the curve and axis, use the disk method.
Rotation about the x-axis:
For y = f(x) rotated about the x-axis on [a, b]:
V = π∫[a to b][f(x)]²dx
The radius of each disk is R(x) = f(x), and the area of each circular cross-section is πR².
Rotation about the y-axis:
For x = g(y) rotated about the y-axis on [c, d]:
V = π∫[c to d][g(y)]²dy
Rotation about other horizontal lines (y = k):
V = π∫[a to b][f(x) - k]²dx
The radius becomes the vertical distance from the curve to the line y = k.
Volumes of revolution — washer method
When there is a gap between the region and the axis of rotation, or when rotating the area between two curves, use the washer method.
General formula:
V = π∫a to b]² - [r(x)]²)dx
where R(x) is the outer radius (distance from axis to outer curve) and r(x) is the inner radius (distance from axis to inner curve).
For rotation about the x-axis:
If the region is bounded by y = f(x) above and y = g(x) below, rotating about the x-axis gives:
V = π∫a to b]² - [g(x)]²)dx
For rotation about y = k:
Calculate distances from the line y = k rather than from y = 0:
V = π∫a to b - k)]² - [(g(x) - k)]²)dx
Always subtract the inner radius squared from the outer radius squared.
Accumulation and net change
The Fundamental Theorem of Calculus establishes that integration accumulates change.
If f'(x) represents a rate of change, then:
∫[a to b]f'(x)dx = f(b) - f(a)
This represents the net change in f from x = a to x = b.
Applications include:
- Displacement: If v(t) is velocity, ∫[t₁ to t₂]v(t)dt gives displacement
- Total distance: ∫[t₁ to t₂]|v(t)|dt gives total distance traveled
- Particle position: If position function is s(t), then s(t) = s(0) + ∫[0 to t]v(x)dx
- Population change: If P'(t) is rate of population change, ∫[t₁ to t₂]P'(t)dt = P(t₂) - P(t₁)
- Water flow: If R(t) is flow rate in liters per hour, ∫[t₁ to t₂]R(t)dt gives total volume
Average value of a function
The average value of a continuous function f on [a, b] is:
f_avg = (1/(b-a))∫[a to b]f(x)dx
This represents the height of a rectangle over [a, b] with the same area as the region under f.
Geometric interpretation:
The average value is the constant height such that the rectangle with base (b - a) and height f_avg has area equal to ∫[a to b]f(x)dx.
Mean Value Theorem for Integrals:
If f is continuous on [a, b], there exists at least one c in (a, b) such that:
f(c) = (1/(b-a))∫[a to b]f(x)dx
This guarantees that f actually attains its average value at some point in the interval.
Worked examples
Example 1: Area between curves
Question: Find the area of the region enclosed by y = x² - 2x and y = x.
Solution:
Step 1: Find intersection points.
Set x² - 2x = x
x² - 3x = 0
x(x - 3) = 0
x = 0 or x = 3
Step 2: Determine which function is greater.
Test x = 1:
- y = x² - 2x = 1 - 2 = -1
- y = x = 1
So y = x is above y = x² - 2x on [0, 3].
Step 3: Set up and evaluate the integral.
Area = ∫[0 to 3](x - (x² - 2x))dx
= ∫[0 to 3](3x - x²)dx
= [(3x²/2) - (x³/3)]₀³
= (27/2 - 9) - 0
= 27/2 - 18/2
= 9/2
Answer: 9/2 or 4.5 square units
Example 2: Volume by washer method
Question: Find the volume of the solid generated by rotating the region bounded by y = √x, y = 0, and x = 4 about the x-axis.
Solution:
Step 1: Identify the method.
The region touches the x-axis, so we use the disk method (special case of washer with inner radius 0).
Step 2: Set up the integral.
Rotating y = √x about the x-axis from x = 0 to x = 4:
V = π∫[0 to 4][√x]²dx
= π∫[0 to 4]x dx
Step 3: Evaluate.
V = π[x²/2]₀⁴
= π(16/2 - 0)
= 8π
Answer: 8π cubic units
Example 3: Accumulation function
Question: A tank contains 200 liters of water at t = 0. Water flows into the tank at rate R(t) = 5 + 2sin(t) liters per minute. How much water is in the tank at t = π minutes?
Solution:
Step 1: Recognize this as a net change problem.
The amount of water added is the integral of the rate.
Step 2: Calculate the accumulation.
Water added = ∫[0 to π](5 + 2sin(t))dt
= [5t - 2cos(t)]₀^π
= (5π - 2cos(π)) - (0 - 2cos(0))
= 5π - 2(-1) - (-2(1))
= 5π + 2 + 2
= 5π + 4
Step 3: Add to initial amount.
Total water = 200 + 5π + 4 = 204 + 5π ≈ 219.7 liters
Answer: (204 + 5π) liters or approximately 219.7 liters
Common mistakes and how to avoid them
Forgetting to subtract functions in correct order: Always use (upper - lower) for vertical slicing or (right - left) for horizontal slicing. Sketch the region to identify which is which.
Using the wrong variable in volumes of revolution: If rotating about the y-axis or integrating with respect to y, express everything in terms of y, not x. Rewrite functions in the form x = g(y).
Confusing disk and washer methods: Check whether there is a gap between the region and the axis of rotation. If rotating the area between two curves, or if the curve doesn't touch the axis, you need the washer method with both outer and inner radii.
Squaring incorrectly in volume problems: Remember that (f(x) - k)² ≠ [f(x)]² - k². Expand properly or keep the expression (f(x) - k)² when setting up the integral.
Ignoring absolute value for total distance: Net displacement uses ∫v(t)dt, but total distance requires ∫|v(t)|dt. Split the integral at points where velocity changes sign.
Mixing up area formulas for cross-sections: Memorize the standard area formulas for common shapes. For an equilateral triangle with side s, the area is (√3/4)s², not (1/2)s².
Exam technique for Applications of Integration
Sketch the region: Even a rough sketch helps identify bounds, intersection points, and which function is greater. On free-response questions, a labeled diagram may earn points.
Show integral setup before evaluating: AP graders award points for correct integral expressions even if arithmetic errors occur later. Write the complete integral with limits and integrand before substituting or simplifying.
Check units and context: In applied problems, state units with your final answer. If the question asks for volume in cubic meters or rate in liters per hour, include these units.
Use your calculator strategically: You may use a calculator on certain sections. For complex integrals or finding intersection points, numerical methods are acceptable and often faster than algebraic manipulation.
Quick revision summary
Applications of integration allow you to calculate areas between curves by integrating (upper - lower), volumes of solids with known cross-sections by integrating area functions, and volumes of revolution using disk method V = π∫R²dx or washer method V = π∫(R² - r²)dx. Net change problems integrate rates to find total accumulation, while average value divides the definite integral by interval length. Always sketch regions, identify correct bounds, and set up integrals carefully before evaluating.