MGU Mathematics Series — Calculus
Part V: Applications of Integration, Differential Equations, and Series
Area · Volume · Work · Mass · Economic Surplus
Chapter 12 of 14
MGU Mathematics Series  ·  Calculus  ·  Library Textbook Edition

Chapter 12:
Applications of
Integration

Area between curves, volumes of revolution, average value, accumulated change, work, mass, and economic surplus — the integral as an accumulation machine applied across geometry, physics, and economics

■ Math Level: Calculus I ■

■ 14 Sections  •  Part V of 5 ■

Follows Chapter 11: Fundamental Theorem of Calculus  ·  Prepares students for Calculus II

Chapter 12 — Front Matter

Purpose and Overview

Chapter 11 revealed the deep unity at the heart of calculus: differentiation and integration are inverse processes, and the Fundamental Theorem of Calculus connects them through the language of accumulation and rate. Now the work shifts. Chapter 12 asks what definite integrals can actually do — not just as mathematical objects, but as tools for measuring real-world quantities.

The answer is: quite a lot. An integral accumulates. Whatever is changing or accumulating over an interval — area, volume, distance, work, mass, economic surplus — a well-constructed integral can measure it. This chapter extends the reach of integration from area under a single curve to area between two curves, from flat regions to three-dimensional solids, from pure mathematics to physics, economics, and engineering.

Each application in this chapter shares a common structure: identify what is being accumulated, write a small representative piece (sometimes called an infinitesimal element or a "slice"), integrate over the relevant interval, and interpret the result in context. That structure — slicing, representing, integrating, interpreting — is the organizing idea of the entire chapter.

Opening Question

A city engineer needs to determine how much steel cable is required to build a suspension bridge. A water district needs to know how much water a reservoir can hold given its cross-sectional shape. A company wants to find the economic benefit its product provides to consumers beyond what they actually paid. An aerospace firm needs to calculate how much work its rocket engine performs lifting a spacecraft out of Earth's atmosphere.

All of these problems involve accumulation — not of a single function, but of a quantity built up from changing contributions over a range. How can a single integral answer questions this different from one another? That is what this chapter explores.

14Sections
11Objectives
14Key Terms
15+Worked Examples

Learning Objectives

  1. 1Set up and evaluate integrals representing the area between two curves.
  2. 2Determine when to integrate with respect to x and when to integrate with respect to y.
  3. 3Compute volumes of solids of revolution using the disk method and washer method.
  4. 4Recognize when the shell method is appropriate and set up shell method integrals.
  5. 5Apply the average value formula to interpret functions in applied contexts.
  6. 6Use integrals to compute accumulated change from rate functions with appropriate units.
  7. 7Set up and evaluate introductory work integrals involving variable force.
  8. 8Set up and evaluate introductory mass integrals using density functions.
  9. 9Set up and evaluate consumer surplus and producer surplus integrals.
  10. 10Identify and correct common errors in integration application problems.
  11. 11Connect integral applications to Calculus II, physics, engineering, and economics.
Chapter 12 — Contents

Sections at a Glance

Sections 12.1 – 12.3
Area: Between Curves and Across Variables
Area Between Curves · Top Minus Bottom · Integrating in x or y
Section 12.1

Area Between Curves

The Idea

Chapter 10 showed that the definite integral ∫ab f(x) dx measures the signed area between the graph of f and the horizontal axis. Chapter 12 extends this idea: instead of measuring area between a curve and the axis, we measure the area between two curves.

Imagine two functions, f(x) and g(x), graphed on the same coordinate plane. Suppose that for every x in [a, b], the graph of f lies above the graph of g. The area of the region sandwiched between the two curves is the area under f minus the area under g.

Area Between Two Curves Area = ∫ab [f(x) − g(x)] dx     (when f(x) ≥ g(x) on [a, b])

Setting Up the Integral

Before computing any integral, ask two questions: (1) Which function is on top over the relevant interval? (2) Where do the curves intersect, and what are the limits of integration? The answer to the first question determines the integrand. The answer to the second question determines [a, b]. Always verify which function is on top by testing a value inside the interval.

Finding Intersection Points

The limits of integration are often the x-coordinates where the two curves intersect. To find these, set f(x) = g(x) and solve for x. Always verify which function is on top by testing a value inside the interval before writing the integrand.

Example 12.1.1
Area Between a Parabola and a Line
  • ProblemFind the area of the region enclosed between f(x) = x² and g(x) = x + 2.
  • ThinkThe region is enclosed — the curves intersect and the area is bounded on both sides. Find the intersections to get the limits, then confirm which function is on top.
  • RepresentSet x² = x + 2: x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = −1 and x = 2 Test at x = 0: f(0) = 0, g(0) = 2. So g(x) = x + 2 is on top over [−1, 2].
  • Calculate Area = ∫₋₁² [(x + 2) − x²] dx = ∫₋₁² (−x² + x + 2) dx = [−x³/3 + x²/2 + 2x]₋₁² At x = 2: −8/3 + 2 + 4 = −8/3 + 6 = 10/3 At x = −1: 1/3 + 1/2 − 2 = 2/6 + 3/6 − 12/6 = −7/6 Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2
  • CheckThe region has width 3 with a maximum gap of about 2 at x = 0. A rough rectangle estimate gives 3 × 2 / 2 = 3. The answer 9/2 = 4.5 is reasonable and slightly larger.
  • InterpretThe area of the region enclosed between the parabola y = x² and the line y = x + 2 is 9/2 square units.
Answer: 9/2 square units
Graph Instruction

Draw a coordinate plane showing the parabola y = x² opening upward and the line y = x + 2. Mark the intersection points at (−1, 1) and (2, 4). Shade the region between the two curves. Label the upper boundary as g(x) = x + 2 and the lower boundary as f(x) = x². Draw a thin vertical rectangle inside the shaded region to illustrate the integrand g(x) − f(x) as the height of the representative slice.

When Curves Switch

If the curves switch which one is on top within the interval, the integral must be split at the crossing point. Compute the area in each piece separately and add the results. Mixing pieces where different functions are on top causes sign errors and gives an incorrect (reduced) total area.

Example 12.1.2
Area Where Curves Cross
  • ProblemFind the total area enclosed between f(x) = sin x and g(x) = cos x on [0, π].
  • ThinkThe curves cross on [0, π]. Find the crossing point and split the integral there.
  • RepresentSet sin x = cos x: tan x = 1, so x = π/4 on [0, π]. On [0, π/4]: cos x is on top (test x = 0: cos 0 = 1 > sin 0 = 0). On [π/4, π]: sin x is on top (test x = π/2: sin(π/2) = 1 > cos(π/2) = 0).
  • Calculate Area = ∫₀^(π/4) [cos x − sin x] dx + ∫_(π/4)^π [sin x − cos x] dx First: [sin x + cos x]₀^(π/4) = (√2/2 + √2/2) − (0 + 1) = √2 − 1 Second: [−cos x − sin x]_(π/4)^π = (−cos π − sin π) − (−cos(π/4) − sin(π/4)) = (1 − 0) − (−√2/2 − √2/2) = 1 + √2 Total = (√2 − 1) + (1 + √2) = 2√2 ≈ 2.83
  • InterpretThe total area enclosed between the sine and cosine curves on [0, π] is 2√2 ≈ 2.83 square units.
Answer: 2√2 square units
Practice 12.1Area Between Curves

For each problem: find intersection points, determine which function is on top, set up and evaluate the integral.

  1. f(x) = x and g(x) = x²
  2. f(x) = 4 − x² and g(x) = x² − 4
  3. f(x) = x³ and g(x) = x
  4. f(x) = √x and g(x) = x²/4 on [0, 4]
  5. f(x) = e^x, g(x) = e, and x = 0 (enclosed region)
  6. f(x) = sin x and g(x) = 0 on [0, 2π] (total area, not net)
Section 12.2

Choosing Top Minus Bottom

The Rule and Its Reason

The phrase "top minus bottom" is the most useful shorthand in area-between-curves problems. When you draw a thin vertical rectangle inside a region, its height is the y-value of the upper boundary minus the y-value of the lower boundary. Integrating those heights over the interval gives the total area. This formula works regardless of whether the region lies above or below the x-axis — even if both curves are below the axis, the difference f(x) − g(x) is still positive when f is above g, and the area comes out correct.

Top Minus Bottom — Always

Area = ∫[top − bottom] dx

Verify which function is on top by testing a point inside the interval. Even when both curves are negative, the formula works. Never guess the order — always check.

Student Warning — Curves That Cross

If the curves cross within the interval and you integrate without splitting, the positive and negative pieces cancel, giving an answer smaller than the true area. Finding the crossing points and splitting the integral is not optional — it is required whenever the top and bottom functions switch.

Example 12.2.1
Region Entirely Below the Axis
  • ProblemFind the area enclosed between f(x) = −x² and g(x) = −4.
  • RepresentSet −x² = −4: x² = 4, so x = ±2. At x = 0: f(0) = 0, g(0) = −4. So f(x) = −x² is on top (less negative).
  • Calculate Area = ∫₋₂² [(−x²) − (−4)] dx = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16/3 + 16/3 = 32/3
  • InterpretThe enclosed area is 32/3 ≈ 10.67 square units, even though both curves lie entirely below the horizontal axis.
Answer: 32/3 ≈ 10.67 square units
Section 12.3

Integrating with Respect to x or y

When to Switch Variables

Most area problems use vertical slices of width dx stacked horizontally. But some regions are described more naturally by their left and right boundaries as functions of y — in those cases, horizontal slices of height dy are the cleaner approach.

Integrating with Respect to y Area = ∫cd [right function − left function] dy
SituationBetter VariableIntegrand
Boundaries are functions of x; each vertical line x = k cuts region cleanly into top and bottomx (vertical slices)top − bottom, dx
Boundaries are functions of y; a single vertical line crosses boundaries multiple timesy (horizontal slices)right − left, dy
Sideways parabola or similar y-dependent curvey (horizontal slices)right − left, dy
Example 12.3.1
Integrating with Respect to y
  • ProblemFind the area enclosed by x = y² and x = y + 2.
  • ThinkBoth boundaries are expressed as functions of y. A vertical line would cross each boundary twice — integrate with respect to y using horizontal slices.
  • RepresentSet y² = y + 2: y² − y − 2 = 0 → (y − 2)(y + 1) = 0. Intersection at y = −1 and y = 2. Test at y = 0: right boundary x = 0 + 2 = 2, left boundary x = 0² = 0. So x = y + 2 is on the right.
  • Calculate Area = ∫₋₁² [(y + 2) − y²] dy = [y²/2 + 2y − y³/3]₋₁² At y = 2: 2 + 4 − 8/3 = 6 − 8/3 = 10/3 At y = −1: 1/2 − 2 + 1/3 = 3/6 − 12/6 + 2/6 = −7/6 Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2
  • InterpretThe area of the region enclosed between x = y² and x = y + 2 is 9/2 square units, computed using horizontal integration.
Answer: 9/2 square units
Graph Instruction

Draw a coordinate plane with the sideways parabola x = y² opening to the right and the line x = y + 2. Mark the intersection points at (1, −1) and (4, 2). Shade the enclosed region. Draw a representative horizontal rectangle of width (y + 2) − y² and height dy to show the integrand for the dy integration.

Practice 12.2–12.3Top Minus Bottom; Variable Choice
  1. Find the area between f(x) = x + 3 and g(x) = x² + 1 on the interval where f is on top. Show the top-minus-bottom setup explicitly.
  2. Find the area of the region enclosed by x = 4 − y² and x = 0. Set up the integral with respect to y.
  3. Find the area enclosed between x = y² − 4 and x = 2 − y² using horizontal slices.
  4. A student computes the area between sin x and cos x on [0, 2π] as a single integral without splitting and gets 0. Explain the geometric error. What is the correct total area?
Sections 12.4 – 12.7
Volume: Slicing, Disks, Washers, and Shells
Volume by Slicing · Disk Method · Washer Method · Shell Method
Section 12.4

Volume by Slicing

The area-between-curves approach built area by adding up infinitely thin vertical strips. Volume can be built the same way — by adding up infinitely thin slices of a three-dimensional solid. If the cross-sectional area of a solid at position x along an axis is A(x), then the volume of the solid is:

Volume by Slicing V = ∫ab A(x) dx

The key step is finding a formula for A(x). Different solids have different cross-sectional shapes. When the cross-section is a square with side length s(x), then A(x) = [s(x)]². When the cross-section is a semicircle with diameter f(x), then A(x) = (π/8)[f(x)]². For circular cross-sections from rotation, the disk and washer methods follow directly.

Example 12.4.1
Volume with Square Cross-Sections
  • ProblemA solid has a base bounded by y = √x and y = 0 on [0, 4]. Every cross-section perpendicular to the x-axis is a square. Find the volume.
  • ThinkThe side length of each square slice equals the height of the base region at position x: s(x) = √x. So A(x) = (√x)² = x.
  • CalculateV = ∫₀⁴ x dx = [x²/2]₀⁴ = 8
  • InterpretThe volume of the solid is 8 cubic units.
Answer: 8 cubic units
Section 12.5

Disk Method

Solids of Revolution

When a region in the plane is rotated around an axis, it sweeps out a three-dimensional solid called a solid of revolution. When the region has no gap between itself and the axis of rotation, each cross-section perpendicular to the axis is a solid disk. The radius of the disk at position x is the function value f(x), and the cross-sectional area is A(x) = π[f(x)]².

Disk Method — Rotation Around the x-axis V = ∫ab π [f(x)]² dx
When to Use the Disk Method

Use the disk method when the region is rotated around the x-axis and the inner edge of the rotation is the axis itself (no hole). Each cross-section should be a solid disk. If there is a gap between the region and the axis, use the washer method instead.

Example 12.5.1
Volume of a Cone Using the Disk Method
  • ProblemFind the volume of the solid formed by rotating f(x) = (r/h)x around the x-axis on [0, h], where r and h are positive constants.
  • ThinkRotating this line segment produces a cone of height h and base radius r. The disk method should reproduce the known formula V = (1/3)πr²h.
  • RepresentAt position x, the disk radius is (r/h)x, so A(x) = π(r/h)²x².
  • Calculate V = ∫₀ʰ π(r/h)²x² dx = (πr²/h²) · [x³/3]₀ʰ = (πr²/h²) · h³/3 = (1/3)πr²h
  • CheckThis matches the familiar geometry formula V = (1/3)πr²h for the volume of a cone. Calculus has reproduced a known result.
  • InterpretThe volume of the cone is (1/3)πr²h.
Answer: V = (1/3)πr²h
Example 12.5.2
Disk Method with an Exponential Function
  • ProblemFind the volume of the solid formed by rotating f(x) = eˣ around the x-axis on [0, 1].
  • Calculate V = ∫₀¹ π[eˣ]² dx = π ∫₀¹ e²ˣ dx = π[e²ˣ/2]₀¹ = (π/2)(e² − 1)
  • InterpretV = (π/2)(e² − 1) ≈ (π/2)(6.389) ≈ 10.04 cubic units.
Answer: (π/2)(e² − 1) ≈ 10.04 cubic units
Section 12.6

Washer Method

When There Is a Hole

If the region being rotated does not touch the axis of rotation — if there is a gap between the region and the axis — then rotating it produces a solid with a hole through the middle. Each cross-section is now a washer: a large disk with a smaller disk removed. For the region between f(x) (upper) and g(x) (lower) rotated around the x-axis, the outer radius is R(x) = f(x) and the inner radius is r(x) = g(x).

Washer Method — Rotation Around the x-axis V = ∫ab π ([f(x)]² − [g(x)]²) dx
Critical Error to Avoid

The washer formula is π([R(x)]² − [r(x)]²), not π[R(x) − r(x)]². These are not the same expression. Squaring first, then subtracting, is correct. Subtracting first, then squaring, is wrong and will give an incorrect answer every time.

Example 12.6.1
Volume Using the Washer Method
  • ProblemFind the volume of the solid formed by rotating the region between f(x) = √x and g(x) = x² around the x-axis on [0, 1].
  • ThinkAt x = 1/4: f(1/4) = 1/2, g(1/4) = 1/16. So f(x) = √x is on top. Outer radius R = √x, inner radius r = x².
  • Calculate V = ∫₀¹ π[(√x)² − (x²)²] dx = π ∫₀¹ (x − x⁴) dx = π[x²/2 − x⁵/5]₀¹ = π(1/2 − 1/5) = π · 3/10 = 3π/10
  • InterpretThe solid formed by rotating the region between √x and x² around the x-axis has volume 3π/10 ≈ 0.94 cubic units.
Answer: 3π/10 ≈ 0.94 cubic units
Graph Instruction

Draw the region bounded above by y = √x and below by y = x² on [0, 1]. Draw a representative horizontal cross-section at position x showing a washer shape with outer radius √x and inner radius x². Label R(x) and r(x). Indicate the axis of rotation along the x-axis and use shading to distinguish the annular cross-section from the hole at the center.

Section 12.7

Shell Method as an Introduction

A Different Way to Slice

The disk and washer methods slice the solid perpendicular to the axis of rotation, producing circular cross-sections. The shell method slices the solid parallel to the axis of rotation, producing thin cylindrical shells — like peeling layers from an onion. When a region bounded above by f(x) and below by the x-axis is rotated around the y-axis, each shell at position x has radius x, height f(x), and thickness dx. Its volume is approximately 2πx · f(x) dx (circumference × height × thickness).

Shell Method — Rotation Around the y-axis V = ∫ab 2π x · f(x) dx
When to Use the Shell Method

Use the shell method when rotating around a vertical axis (like the y-axis) and the function is expressed in terms of x. It avoids the need to solve for x in terms of y, which the disk method would require in this situation.

Example 12.7.1
Shell Method — Rotating Around the y-axis
  • ProblemFind the volume of the solid formed by rotating the region under f(x) = x² on [0, 2] around the y-axis.
  • ThinkRotating around the y-axis. The disk method would require expressing x in terms of y. Use the shell method to keep everything in x.
  • Calculate V = ∫₀² 2π x · x² dx = 2π ∫₀² x³ dx = 2π[x⁴/4]₀² = 2π · 4 = 8π
  • InterpretThe volume of the solid of revolution is 8π ≈ 25.13 cubic units.
Answer: 8π ≈ 25.13 cubic units
MGU Library Connection
  • Shell method: Calculus II, extended volume problems with non-standard axes of rotation
  • Moments of inertia and centroids using shells: MGU Physics Library — rotational dynamics
  • Solid mechanics and material cross-sections: MGU Engineering Mathematics pages
Practice 12.4–12.7Volume Problems

For each problem, identify the method (disk, washer, or shell), set up the integral, and evaluate.

  1. f(x) = √x around x-axis on [0, 9]
  2. f(x) = x², g(x) = x; washer around x-axis on [0, 1]
  3. f(x) = 2 − x² around x-axis on [−1, 1]
  4. f(x) = x² around y-axis on [0, 3] (shell method)
  5. f(x) = √(r² − x²) around x-axis on [−r, r]; confirm V = (4/3)πr³
  6. f(x) = cos x around x-axis on [0, π/2]
  7. Shell: f(x) = 4x − x² around y-axis on [0, 4]
  8. f(x) = eˣ, x = 0, x = 1, y = 0; disk around x-axis
Sections 12.8 – 12.12
Physics, Economics, and Accumulated Change
Average Value · Accumulated Change · Work · Mass · Surplus
Section 12.8

Average Value in Applications

The average of a finite list of numbers sums them and divides by the count. For a function that takes infinitely many values over an interval, the definite integral provides the continuous analog: accumulate the function values and divide by the length of the interval.

Average Value of a Function on [a, b] favg = (1/(b − a)) ∫ab f(x) dx

Average value is meaningful whenever a function varies continuously over a range and a single representative value is needed: temperature over a day, speed over a trip, profit rate over a quarter, voltage over a cycle. Geometrically, the average value is the height of the rectangle on [a, b] whose area equals the area under the curve.

Example 12.8.1
Average Temperature Over a Day
  • ProblemThe temperature on a summer day is T(t) = 60 + 20 sin(πt/12) °F, where t is hours after midnight. Find the average temperature over a full 24-hour day.
  • Calculate T_avg = (1/24) ∫₀²⁴ [60 + 20 sin(πt/12)] dt = (1/24) [60t − 20 · (12/π) cos(πt/12)]₀²⁴ = (1/24) [(1440 − (240/π)cos(2π)) − (0 − (240/π)cos(0))] = (1/24) [1440 − 240/π + 240/π] = (1/24)(1440) = 60 °F
  • InterpretThe average temperature is exactly 60°F — the baseline of the model. This makes sense because sin averages to zero over a full period, leaving only the constant term.
Answer: 60°F
Example 12.8.2
Average Velocity
  • ProblemA car's velocity is v(t) = 3t² − 6t + 4 m/s on [0, 3]. Find the average velocity.
  • Calculate v_avg = (1/3) ∫₀³ (3t² − 6t + 4) dt = (1/3)[t³ − 3t² + 4t]₀³ = (1/3)[(27 − 27 + 12) − 0] = 12/3 = 4 m/s
  • InterpretThe average velocity is 4 m/s — the constant speed that would produce the same total displacement over the three-second interval.
Answer: 4 m/s
Section 12.9

Accumulated Change from Rates

One of the most practical applications of integration: if a quantity is changing at a known rate, the total accumulated change over a time interval is the definite integral of that rate. Units guide every calculation — if the rate has units of liters per minute and time is in minutes, the integral has units of liters.

Total Accumulated Change Total change = ∫ab rate(t) dt
Example 12.9.1
Water Flow Into a Tank
  • ProblemWater flows into a tank at r(t) = 4t + 3 liters/minute. How much water accumulates from t = 0 to t = 5 minutes?
  • Calculate∫₀⁵ (4t + 3) dt = [2t² + 3t]₀⁵ = (50 + 15) − 0 = 65 liters
  • InterpretThe tank accumulates 65 liters of water during the five-minute interval.
Answer: 65 liters
Example 12.9.2
Population Growth
  • ProblemA bacterial colony grows at P'(t) = 300e^(0.2t) bacteria/hour. How much does the population increase from t = 0 to t = 4 hours?
  • Calculate ∫₀⁴ 300e^(0.2t) dt = 300 · (1/0.2)[e^(0.2t)]₀⁴ = 1500(e^0.8 − 1) ≈ 1500(2.2255 − 1) = 1500(1.2255) ≈ 1838 bacteria
  • InterpretThe bacterial population increases by approximately 1838 bacteria over four hours.
Answer: ≈ 1838 bacteria
Section 12.10

Work as an Introduction

In physics, work is the energy transferred when a force moves an object through a displacement. When force is constant, work = force × distance. When force varies with position, calculus is needed.

Work with Variable Force W = ∫ab F(x) dx

Units: if force is in Newtons and distance in meters, work is in joules (J). If force is in pounds and distance in feet, work is in foot-pounds (ft·lb). Think of the work integral as accumulating tiny contributions: at each position x, the force F(x) acts through an infinitesimally small displacement dx, doing work F(x) dx. Integrating sums all these contributions.

Example 12.10.1
Stretching a Spring (Hooke's Law)
  • ProblemA spring obeys Hooke's Law: F(x) = kx, where x is the stretch distance. If k = 200 N/m, how much work is required to stretch the spring 0.3 m?
  • CalculateW = ∫₀^0.3 200x dx = 200[x²/2]₀^0.3 = 100(0.09) = 9 joules
  • InterpretStretching the spring 30 cm requires 9 joules of work. Note that using the maximum force (200 × 0.3 = 60 N) times the distance would give 18 J — twice as much — because the force starts at zero and builds up. Integration correctly averages the variable force.
Answer: 9 joules
Example 12.10.2
Pumping Water Out of a Cylindrical Tank
  • ProblemA cylindrical tank has radius 2 m and height 5 m and is full of water (density 1000 kg/m³, g = 9.8 m/s²). How much work is required to pump all the water out over the top?
  • ThinkA thin horizontal slice at height y must be lifted a distance (5 − y) to the top. The volume of the slice is π(2²) dy = 4π dy. Its weight is 1000 · 9.8 · 4π dy Newtons.
  • Calculate W = ∫₀⁵ 1000 · 9.8 · 4π(5 − y) dy = 39,200π ∫₀⁵ (5 − y) dy = 39,200π [5y − y²/2]₀⁵ = 39,200π · 25/2 = 490,000π ≈ 1,539,380 joules ≈ 1.54 MJ
  • InterpretPumping the full tank dry requires approximately 1.54 megajoules — a concrete illustration of why large tanks require powerful pumps. Water near the top is easy to lift; water near the bottom must travel the full 5 meters.
Answer: 490,000π ≈ 1.54 MJ
MGU Library Connection
  • Work, energy, and power: MGU Physics Library — mechanics and thermodynamics
  • Pumping problems with non-cylindrical tanks: MGU Engineering Mathematics pages
  • Variable force and impulse: MGU Physics Library — dynamics chapter
Section 12.11

Mass and Density as an Introduction

When an object has uniform density, mass = density × volume (or length). But many real objects have density that varies with position. When density varies along a single dimension, the mass of a thin slice at position x with thickness dx is dm = ρ(x) dx, where ρ(x) is the linear density function (mass per unit length). Integrating gives the total mass.

Mass with Variable Linear Density m = ∫ab ρ(x) dx
Example 12.11.1
Mass of a Rod with Variable Density
  • ProblemA metal rod extends from x = 0 to x = 3 m. Its linear density is ρ(x) = 2 + x kg/m. Find the total mass.
  • Calculate m = ∫₀³ (2 + x) dx = [2x + x²/2]₀³ = 6 + 9/2 = 21/2 = 10.5 kg
  • InterpretThe rod has a total mass of 10.5 kg. If the density were constant at 2 kg/m, the mass would be 6 kg. The additional 4.5 kg comes from the increasing density toward the rod's far end.
Answer: 10.5 kg
Extension to Two Dimensions

When density varies over a two-dimensional region, mass is computed with a double integral — a topic in Calculus III. For now, one-dimensional density problems are the focus. The same slicing idea extends naturally: if a thin plate has area density σ(x) varying in one direction and width w(x), then m = ∫ σ(x) · w(x) dx.

Section 12.12

Consumer and Producer Surplus as an Introduction

Market Equilibrium

In economics, a demand function D(x) gives the price consumers are willing to pay for x units of a good. A supply function S(x) gives the price at which producers are willing to supply x units. At equilibrium, supply equals demand at quantity x* and price p* = D(x*) = S(x*).

ConceptMeaningFormula
Consumer SurplusBenefit to buyers who would have paid more than p*CS = ∫₀^(x*) [D(x) − p*] dx
Producer SurplusBenefit to sellers who would have accepted less than p*PS = ∫₀^(x*) [p* − S(x)] dx
Total SurplusCombined market benefit at equilibriumCS + PS
Example 12.12.1
Consumer and Producer Surplus
  • ProblemDemand: D(x) = 100 − 2x dollars. Supply: S(x) = 20 + 3x dollars. Find equilibrium and compute both surpluses.
  • RepresentSet D(x) = S(x): 100 − 2x = 20 + 3x → 80 = 5x → x* = 16. Price: p* = 100 − 32 = 68 dollars.
  • Calculate CS CS = ∫₀¹⁶ [(100 − 2x) − 68] dx = ∫₀¹⁶ (32 − 2x) dx = [32x − x²]₀¹⁶ = (512 − 256) − 0 = 256 dollars
  • Calculate PS PS = ∫₀¹⁶ [68 − (20 + 3x)] dx = ∫₀¹⁶ (48 − 3x) dx = [48x − 3x²/2]₀¹⁶ = (768 − 384) − 0 = 384 dollars
  • InterpretConsumers gain $256 in surplus — the benefit of paying $68 instead of the higher prices they would have accepted. Producers gain $384 in surplus. Total market surplus is $640.
Consumer Surplus: $256  ·  Producer Surplus: $384  ·  Total: $640
Graph Instruction

Draw a graph with quantity x on the horizontal axis and price p on the vertical. Draw a downward-sloping demand curve and an upward-sloping supply curve. Mark the equilibrium point at (16, 68). Shade consumer surplus as the triangle above the horizontal line p = 68 and below the demand curve. Shade producer surplus as the triangle below p = 68 and above the supply curve. Label both shaded regions clearly.

MGU Library Connection
  • Welfare economics and market efficiency: MGU Economics Library — consumer and producer theory
  • Integration in economic modeling: MGU Data Science Foundations — economic modeling chapter
  • Gini coefficient and income distributions: MGU Statistics Library — inequality measures
Practice 12.8–12.12Average Value, Rates, Work, Mass, and Surplus
  1. Average Value. Find the average value of f(x) = x² + 1 on [0, 3].
  2. Accumulated Change. A population grows at P'(t) = 500e^(0.05t) people per year. How much does the population increase over 10 years?
  3. Accumulated Change. Water drains at r(t) = 8 − 0.4t liters/min. How much drains in the first 10 minutes?
  4. Work. A spring has k = 150 N/m. How much work is done compressing the spring 0.4 m from its natural length?
  5. Work. A cable weighing 5 kg/m hangs 20 m down a shaft. How much work is done pulling the entire cable to the top? (Hint: as the cable is pulled up, only the remaining length needs lifting.)
  6. Mass. A rod from x = 0 to x = 4 m has density ρ(x) = 3 + 2√x kg/m. Find the total mass.
  7. Surplus. Demand: D(x) = 80 − 5x. Supply: S(x) = 20 + 3x. Find equilibrium, consumer surplus, and producer surplus.
  8. Average Power. P(t) = 400 + 100 cos(πt/6) watts. Find the average power output over 12 hours.
Sections 12.13–12.14 and Back Matter

Common Mistakes, Preparation, and Review

12.13 — Common Integration Application Mistakes

Mistake 1 — Forgetting to Subtract for Area Between Curves
Area = ∫f(x) dx − ∫g(x) dx (computed separately, sign errors follow)

Correct approach: always write ∫[f(x) − g(x)] dx with the top function first. Computing the integrals separately and subtracting risks sign errors, especially when g is negative. Write the difference inside a single integral.

Mistake 2 — Not Splitting When Curves Cross
∫₀^π [sin x − cos x] dx = 0 (positive and negative areas cancel)

Correct: find the crossing point (x = π/4), split the integral, and add the absolute values. Reporting a near-zero answer as the "area" when the region clearly covers more than that should trigger a check for a crossing.

Mistake 3 — Using the Disk Formula When There Is a Hole
V = ∫π[f(x)]² dx (when g(x) > 0 creates an inner radius)

Correct: use the washer formula V = ∫π([f(x)]² − [g(x)]²) dx. The disk method counts the hole as solid material, giving a volume that is too large.

Mistake 4 — Squaring the Difference Instead of Subtracting Squares
V = ∫π[f(x) − g(x)]² dx (subtracted first, then squared)

Correct: V = ∫π([f(x)]² − [g(x)]²) dx. These are not equal: [f − g]² = f² − 2fg + g², which includes the cross term −2fg and is almost never equal to f² − g².

Mistake 5 — Ignoring Units in Work and Mass Problems
W = ∫F(x) dx = 9 (no units stated)

Correct: W = 9 joules (or ft·lb, depending on unit system). Units are not optional in applied problems. They confirm that the integral is measuring what it is supposed to measure and catch errors in the setup.

Mistake 6 — Wrong Shell Method Radius
V = ∫2π x · f(x) dx when rotating around x = 3 (uses x instead of |x − 3|)

Correct: when rotating around the line x = c, the shell radius is |x − c|, not x. For rotation around the y-axis, x is correct. For other vertical axes, adjust the radius expression accordingly.

Mistake 7 — Guessing Which Function Is on Top
Assuming f(x) = x² is always below g(x) = 2x without testing

Correct: always test a value in the interior of the interval. Functions can switch which is on top at different parts of the domain. Testing takes five seconds and prevents sign errors.

Mistake 8 — Confusing Net Area with Total Area
Area = ∫₋₁¹ (x³ − x) dx = 0 (net area reported as total area)

Correct: find where the integrand changes sign, split, and add the absolute values of each piece. Net area can be zero even when the total area between the curve and the axis is substantial. Geometric area is always non-negative.

12.14 — Preparing for Calculus II

Chapter 12 closes the first calculus course by extending the reach of the definite integral across geometry, physics, and economics. Every application followed the same organizing idea: identify what is being accumulated, write a representative element, integrate over the relevant interval, and interpret the result in context.

Calculus II typically begins where this chapter ends. Integration techniques — integration by parts, trigonometric integrals, trigonometric substitution, partial fractions, and improper integrals — vastly expand the range of integrands that can be handled. Arc length and surface area of revolution extend the geometric applications. Fluid pressure and center of mass use the same slicing logic applied to new physical contexts.

The integral is not a single formula — it is a way of thinking about accumulated structure. The disk, washer, and shell methods are three manifestations of the same underlying idea. The work, mass, and surplus integrals are three more. In Calculus II, III, differential equations, physics, and economics, this thinking will continue to grow and ramify in every quantitative direction students pursue.

Chapter Summary

Key Terms Review

area between curves
The integral of (top function − bottom function) over an interval, giving the net area of the bounded region between the two graphs.
solid of revolution
A three-dimensional solid formed by rotating a planar region around a line called the axis of rotation.
disk method
A technique for finding volume by integrating circular cross-sectional areas: V = ∫π[f(x)]² dx. Used when there is no hole through the solid.
washer method
A technique for finding volume when the solid has a hole, using annular (washer-shaped) cross-sections: V = ∫π([R(x)]² − [r(x)]²) dx.
shell method
A technique for finding volume using cylindrical shells parallel to the axis of rotation: V = ∫2πx · f(x) dx. Often used for rotation around a vertical axis.
representative slice
A thin element (rectangle, disk, washer, or shell) that, when integrated, builds up the total area, volume, work, or mass of the object.
average value
The continuous analog of averaging: f_avg = (1/(b−a)) ∫f(x) dx over [a, b]. The height of the equal-area rectangle on the interval.
accumulated change
The total change in a quantity computed by integrating its rate function: total change = ∫rate(t) dt.
work
The integral of a variable force over a displacement: W = ∫F(x) dx. Measured in joules (N·m) or foot-pounds.
linear density
A function ρ(x) giving mass per unit length at position x along a rod or wire. Mass = ∫ρ(x) dx.
consumer surplus
The area between the demand curve and the equilibrium price: ∫[D(x) − p*] dx. Represents the benefit consumers receive beyond what they paid.
producer surplus
The area between the equilibrium price and the supply curve: ∫[p* − S(x)] dx. Represents the extra benefit producers receive above their minimum acceptable price.
axis of rotation
The line around which a planar region is rotated to form a solid of revolution.
net area vs. total area
Net area accounts for sign (positive above, negative below a reference). Total area always adds absolute values. The definite integral computes net area; geometric area requires splitting at sign changes.

Concept Review Questions

  1. Explain in your own words why the area between two curves is computed as the integral of (top minus bottom). Why can't you just integrate each function separately and subtract?
  2. When would you choose to integrate with respect to y instead of x? Give a specific example of a region where the dy approach is cleaner.
  3. Describe the difference between the disk method and the washer method. What feature of the solid determines which method to use?
  4. The shell method integrates parallel to the axis of rotation while the disk method integrates perpendicular to it. Describe the shape of a representative element in each method.
  5. A function's average value on [0, 4] is 7. What does this mean geometrically? Can you sketch a possible function?
  6. A rate function r(t) is measured in gallons per hour. What does ∫₂⁵ r(t) dt represent, and what are its units?
  7. Why does computing work with a variable force require integration? What happens to the formula when force is constant?
  8. What is consumer surplus, and why does it involve integration rather than simple multiplication by the equilibrium quantity?
Practice AArea Between Curves

For each problem: (a) find intersection points, (b) determine which function is on top, (c) set up the integral, (d) evaluate and state the area with units.

  1. f(x) = x and g(x) = x²
  2. f(x) = 4 − x² and g(x) = x² − 4
  3. f(x) = sin x on [0, 2π], g(x) = 0 (total area, not net)
  4. f(x) = x³ and g(x) = x
  5. x = y² − 4 and x = 2 − y² (horizontal slices)
  6. f(x) = eˣ, g(x) = e, x = 0 (find the enclosed region)
  7. f(x) = x³ − x and g(x) = 0 on [−1, 1] (total area)
  8. Set up (do not evaluate): f(x) = ln x, y = 0, x = e²
Practice BVolume

Identify the method, set up the integral, and evaluate.

  1. f(x) = √x around x-axis on [0, 9]
  2. f(x) = x², g(x) = x; washer around x-axis on [0, 1]
  3. f(x) = 2 − x² around x-axis on [−1, 1]
  4. Shell: f(x) = x² around y-axis on [0, 3]
  5. f(x) = √(r² − x²) around x-axis on [−r, r]; confirm V = (4/3)πr³
  6. f(x) = eˣ around x-axis on [0, 1]
  7. f(x) = cos x around x-axis on [0, π/2]
  8. Shell: f(x) = 4x − x² around y-axis on [0, 4]
Practice CAccumulated Change Problems
  1. A population grows at P'(t) = 500e^(0.05t) per year. How much does it increase over 10 years?
  2. Water drains at r(t) = 8 − 0.4t liters/minute. How much drains in the first 10 minutes?
  3. A factory emits E(t) = 200 + 20 sin(πt/6) tons of CO₂ per month. Find total emissions from t = 0 to t = 12 months.
  4. Revenue grows at R'(t) = 1000√t dollars/month. Find total revenue from month 4 to month 9.
  5. A car has acceleration a(t) = 6t m/s². Find the change in velocity and the total distance traveled over the first 4 seconds, starting from rest.
Practice DPhysical and Economic Applications
  1. Spring Work. A spring has k = 150 N/m. How much work is done compressing the spring 0.4 m from its natural length?
  2. Pumping Work. A hemispherical bowl of radius 3 m is full of water (density 1000 kg/m³). Set up the integral to compute the work required to pump all the water over the rim. (Hint: the water surface at height y has horizontal cross-sectional radius √(9 − y²).)
  3. Variable-Density Rod. A rod from x = 0 to x = 4 m has linear density ρ(x) = 3 + 2√x kg/m. Find the total mass.
  4. Consumer and Producer Surplus. Demand: D(x) = 80 − 5x. Supply: S(x) = 20 + 3x. Find equilibrium, consumer surplus, and producer surplus.
  5. Average Power. P(t) = 400 + 100 cos(πt/6) watts. Find average power over 12 hours.
  6. Total Surplus. Demand: D(x) = 120 − 4x. Supply: S(x) = 10 + 6x. Find total surplus (CS + PS) at equilibrium.
  7. Cable Work. A cable weighing 5 kg/m hangs from the top of a 20-meter shaft. How much work is done lifting the entire cable to the top? (Only the remaining hanging portion must be lifted at each moment.)

Error Analysis

Each problem contains an error. Identify and explain the error, then provide the correct setup or solution.

Problem 1
Area between f(x) = x² and g(x) = 2x on [0, 2]: Area = ∫₀² x² dx − ∫₀² 2x dx = 8/3 − 4 = −4/3 Student reports area = 4/3 square units.

The arithmetic is correct but the setup may be wrong. Which function is on top over [0, 2]? Rewrite with the correct function first and verify the answer changes.

Problem 2
Volume of solid from f(x) = 3 and g(x) = x around x-axis on [0, 3]: Student uses V = ∫₀³ π(3 − x)² dx (disk formula with the difference)

Identify the error. Should the washer method be used here? Write the correct integral and explain what the student's setup computes geometrically.

Problem 3
Average value of f(x) = x² on [0, 3]: Student writes f_avg = ∫₀³ x² dx = [x³/3]₀³ = 9 and reports the average value is 9.

What step did the student forget? Apply the correct formula and find the true average value.

Problem 4
Consumer surplus for D(x) = 50 − 2x, equilibrium at x* = 10, p* = 30: CS = (50 − 30) × 10 = 200 (rectangle, no integration)

The student computed a rectangle. What geometric shape is the consumer surplus actually? Set up and evaluate the correct integral.

Problem 5
Work stretching spring (k = 100 N/m) from 0 to 0.5 m: W = F × d = (100 × 0.5) × 0.5 = 25 J (used final force as constant)

The force is not constant — it ranges from 0 to 50 N. Set up and evaluate the correct integral. Why does using the final force double the answer?

Chapter 12 Checkpoint

Chapter 12 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 12
Section 1 — Area Between Curves
  1. Find the area of the region enclosed between f(x) = 6x − x² and g(x) = x. (a) Find the intersection points. (b) Determine which function is on top. (c) Set up and evaluate the integral. (d) State the area with units.
  2. Find the area of the region bounded by x = 4 − y² and x = 0 using horizontal slices. Set up the integral with respect to y, evaluate, and state the area.
Section 2 — Volume
  1. Consider the region bounded by f(x) = √x and g(x) = x/2 on [0, 4]. (a) Verify which function is on top and find the intersection points. (b) Compute the volume of the solid formed by rotating this region around the x-axis using the washer method. (c) Show the setup clearly and evaluate the integral.
  2. Find the volume of the solid formed by rotating the region under f(x) = x³ on [0, 2] around the y-axis using the shell method.
Section 3 — Average Value and Accumulated Change
  1. The temperature in a warehouse during an 8-hour workday is T(t) = 65 + 5t − t²/4 °F, 0 ≤ t ≤ 8. (a) Find the average temperature over the workday. (b) At what time during the day does the temperature equal the average?
  2. A river's flow rate is r(t) = 200 + 30t − 5t² cubic meters per hour, where t is hours since sunrise. Find the total volume of water that flows past a measuring point from t = 0 to t = 6 hours. Interpret the answer with units.
Section 4 — Work and Economic Applications
  1. A spring has k = 80 N/m. How much work is required to stretch the spring from 0.1 m to 0.5 m beyond its natural length?
  2. The demand function is D(x) = 60 − 3x and the supply function is S(x) = 15 + 2x, where x is in hundreds of units and price is in dollars. (a) Find the equilibrium quantity and price. (b) Compute consumer surplus. (c) Compute producer surplus. (d) Interpret each surplus in one sentence.
Section 5 — Error Analysis
  1. A student computes the volume of the solid formed by rotating the region between f(x) = x + 1 and g(x) = x²/2 around the x-axis on [0, 2] using V = ∫₀² π[(x + 1) − x²/2]² dx. Identify the error and write the correct integral.
  2. A student computes the area between f(x) = cos x and g(x) = sin x on [0, 2π] as a single integral and reports an area of 0. Explain the geometric error and describe the correct procedure.
Bridge to Calculus II and Beyond
What Integration Becomes in Higher Mathematics

Chapter 12 closes the first calculus course in the MGU Mathematics Series. Every application followed the same organizing principle: break the quantity into representative pieces, write a function that describes each piece, and integrate over the interval. That principle will never stop being useful.

In Calculus II, integration techniques — integration by parts, trigonometric integrals, partial fractions, and improper integrals — vastly expand the class of functions that can be evaluated in closed form. Arc length and surface area of revolution extend the geometric work. Sequences and series open an entirely new world of infinite-process mathematics.

In Calculus III, double and triple integrals over two- and three-dimensional regions make it possible to compute volumes, masses, centers of mass, and moments of inertia for complex solids. In Differential Equations, the integral is the primary tool for recovering a function from its rate of change. In Physics and Engineering, every major quantity — work, flux, circulation, pressure, potential — is defined by an integral over a region, surface, or path.

The integral is not a topic in calculus. It is a language. Chapter 12 is where students begin to speak it fluently.

MGU Library Connections — Chapter 12
  • Volume formulas reference: Appendix N of this textbook (Area and Volume by Integration)
  • Common Integrals Reference: Appendix K of this textbook
  • Extended integration techniques: Calculus II, Chapters 1–4 (Parts, Trig Integrals, Partial Fractions, Improper)
  • Double and triple integrals: Calculus III, Chapters 3–5
  • Work, energy, fluid pressure, and center of mass: MGU Physics Library — mechanics chapters
  • Consumer/producer surplus, Gini coefficient, and income distributions: MGU Economics Library and MGU Data Science Foundations
  • Moments of inertia and rotating bodies: MGU Engineering Mathematics — solid mechanics