Applications of Integrals

Introduction

Integrals turn local information into accumulated quantities. In this aiMOOC for Grades 11–13, you will learn how a definite integral can represent signed area, geometric area, volume, average value, displacement, distance, work, arc length, and probability. The central habit is not simply to calculate an antiderivative: you must first decide what is being accumulated, choose a suitable variable, determine the bounds, and interpret the result with correct units.

You should already be comfortable with functions, graphs, basic derivatives, and common antiderivatives. A graphing calculator or dynamic mathematics tool can help you check intersections and numerical values, but every model should also be explained in words.

Datei:Integral example.svg

The picture above emphasizes a key idea: a definite integral gives signed accumulation. Contributions above the horizontal axis are positive and contributions below it are negative. Geometric area, by contrast, is never negative, so intervals may need to be split or an absolute value may be required.

Learning Goals

By the end of the course, you should be able to translate a real or geometric situation into an integral, justify the integrand and bounds, evaluate or approximate the integral, attach meaningful units, and check whether the result is reasonable. You should also be able to compare different integration models and explain why two superficially similar problems may require different formulas.

From Sums to Accumulation

A definite integral is the limit of sums of small contributions. If an interval from $a$ to $b$ is divided into many narrow pieces of width $Δ x$ , a quantity such as area can be approximated by a Riemann sum $\sum f (x_{i}^{*}) Δ x .$ As the largest subinterval width tends to zero, the sum approaches $\int_{a}^{b} f (x) d x .$

This viewpoint matters in applications because the factor $f (x_{i}^{*}) Δ x$ usually has a direct meaning: height times width for area, cross-sectional area times thickness for volume, velocity times time for displacement, or force times distance for work.

Datei:Riemann sum convergence.svg

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects accumulation with antiderivatives. If $F^{'} (x) = f (x)$ and $f$ is continuous on the interval, then $\int_{a}^{b} f (x) d x = F (b) - F (a) .$ This theorem lets you evaluate many accumulation models exactly after the model has been set up correctly. In applications, the difficult step is often not integration itself but deciding which quantity belongs inside the integral.

Application 1: Signed Area and Geometric Area

For a continuous function $f$ , the definite integral $\int_{a}^{b} f (x) d x$ is the net signed area between the graph and the horizontal axis. If $f (x)$ is positive on the whole interval, this equals ordinary geometric area. If the graph crosses the axis, the negative contribution subtracts from the positive contribution.

To find total geometric area between a graph and the horizontal axis, use $\int_{a}^{b} | f (x) | d x$ or split the interval at each zero and make each piece positive.

Datei:Integral-area-under-curve.svg

Area Between Two Curves

Suppose $f (x) \geq g (x)$ on $[a, b]$ . A thin vertical strip has approximate area $[f (x) - g (x)] Δ x .$ Therefore the area between the curves is $A = \int_{a}^{b} [f (x) - g (x)] d x .$

The practical rule is top minus bottom for vertical slices. For horizontal slices, rewrite the boundaries as functions of $y$ and use right minus left.

Worked example. Find the area enclosed by $y = 2 x$ and $y = x^{2}$ . The curves intersect when $2 x = x^{2}$ , so $x = 0$ and $x = 2$ . On this interval, $2 x$ is above $x^{2}$ . Thus $A = \int_{0}^{2} (2 x - x^{2}) d x = [x^{2} - \frac{x^{3}}{3}]_{0}^{2} = \frac{4}{3} .$ The area is $\frac{4}{3}$ square units.

Application 2: Volumes from Cross Sections

If a solid extends from $x = a$ to $x = b$ and its cross-sectional area perpendicular to the $x$ -axis is $A (x)$ , then $V = \int_{a}^{b} A (x) d x .$ The logic is area times a small thickness: $A (x) Δ x$ approximates the volume of one thin slab.

Worked example. A solid has a base on $0 \leq x \leq 4$ , and each cross section perpendicular to the $x$ -axis is a square with side length $\sqrt{x}$ . Then $A (x) = (\sqrt{x})^{2} = x$ , so $V = \int_{0}^{4} x d x = 8$ cubic units.

Solids of Revolution: Disk Method

For rotation about a horizontal axis, slices perpendicular to the axis can become disks. If the radius from the axis of rotation to the curve is $R (x)$ , then a disk has area $π R (x)^{2}$ , giving $V = π \int_{a}^{b} R (x)^{2} d x .$

Datei:Generating fx disk method.svg

Worked example. Rotate the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$ -axis. Here $R (x) = \sqrt{x}$ , so $V = π \int_{0}^{4} x d x = 8 π$ cubic units.

Solids of Revolution: Washer Method

If rotation creates a hole, subtract the inner disk from the outer disk. For slices perpendicular to a horizontal axis, with outer radius $R (x)$ and inner radius $r (x)$ , $V = π \int_{a}^{b} (R (x)^{2} - r (x)^{2}) d x .$ The radii are distances from the axis of rotation, so you should draw the axis and label both distances before writing the integral.

Datei:Generating fxs washer method.svg

Worked example. Rotate the region between $y = \sqrt{x}$ and $y = x / 2$ from $x = 0$ to $x = 4$ about the $x$ -axis. The outer radius is $\sqrt{x}$ and the inner radius is $x / 2$ . Therefore $V = π \int_{0}^{4} (x - \frac{x^{2}}{4}) d x = \frac{8 π}{3}$ cubic units.

Application 3: Average Value of a Function

For a continuous function $f$ on $[a, b]$ , its average value is $f_{a v g} = \frac{1}{b - a} \int_{a}^{b} f (x) d x .$ Geometrically, this is the height of a rectangle of width $b - a$ that has the same signed area as the region represented by the integral.

Worked example. For $f (x) = x^{2}$ on $[0, 3]$ , $f_{a v g} = \frac{1}{3} \int_{0}^{3} x^{2} d x = \frac{1}{3} \cdot 9 = 3.$ If $f$ represents temperature, concentration, power, or another continuously varying quantity, this formula gives its time- or space-average over the interval.

Application 4: Motion from Velocity

If $v (t)$ is velocity, then $\int_{a}^{b} v (t) d t$ gives displacement, the net change in position. Total distance traveled is $\int_{a}^{b} | v (t) | d t$ . This distinction is essential when velocity changes sign.

Fehler beim Erstellen des Vorschaubildes:

Worked example. Let $v (t) = 2 t - 4$ for $0 \leq t \leq 4$ . The velocity is zero at $t = 2$ . The displacement is $\int_{0}^{4} (2 t - 4) d t = 0.$ The object returns to its starting position. The total distance is $\int_{0}^{2} (4 - 2 t) d t + \int_{2}^{4} (2 t - 4) d t = 4 + 4 = 8$ distance units.

Application 5: Work by a Variable Force

In one-dimensional motion, if a force $F (x)$ acts in the direction of motion from $x = a$ to $x = b$ , the work done is $W = \int_{a}^{b} F (x) d x .$ A small contribution $F (x) Δ x$ has units of force times distance, so in SI units the integral gives joules.

Datei:Force-distance-diagram.svg

Worked example. If $F (x) = 3 x^{2}$ newtons for $0 \leq x \leq 2$ metres, then $W = \int_{0}^{2} 3 x^{2} d x = [x^{3}]_{0}^{2} = 8$ joules.

Application 6: Arc Length

Integration can measure distance along a curved path. If $y = f (x)$ is continuously differentiable on $[a, b]$ , then its arc length is $L = \int_{a}^{b} \sqrt{1 + {(f^{'} (x))}^{2}} d x .$ The formula comes from applying the Pythagorean theorem to many tiny line segments and passing to a limit.

Datei:Arclength.svg

Worked example. Let $y = \frac{2}{3} x^{3 / 2}$ on $[0, 3]$ . Then $y^{'} = \sqrt{x}$ , so $L = \int_{0}^{3} \sqrt{1 + x} d x = {[\frac{2}{3} (1 + x)^{3 / 2}]}_{0}^{3} = \frac{14}{3}$ length units.

Application 7: Probability from a Density

For a continuous random variable with probability density function $f (x)$ , probability over an interval is accumulated area: $P (a \leq X \leq b) = \int_{a}^{b} f (x) d x .$ A valid density is nonnegative and has total area $1$ over its full domain.

Datei:Valid-probability-density-function.svg

Worked example. Suppose $f (x) = 2 x$ for $0 \leq x \leq 1$ and $f (x) = 0$ elsewhere. Then $P (0 \leq X \leq 0.5) = \int_{0}^{0.5} 2 x d x = 0.25.$

Choosing and Checking an Integral Model

Before calculating, ask what a thin slice represents. For area it is height times width; for volume it is cross-sectional area times thickness; for displacement it is velocity times time; for work it is force times displacement. Then check the bounds, sign, and units.

A strong solution normally follows this sequence:

Model the quantity: State what is accumulating and choose the variable of integration.
Sketch or inspect the graph: Find intersections, zeros, and the relative positions of curves.
Build the integral: Write the correct integrand and bounds before evaluating.
Check units: Area should have square units, volume cubic units, and work force times distance.
Check reasonableness: Compare with a rough geometric or numerical estimate.

Common errors include treating signed area as geometric area, reversing top and bottom curves, using a diameter where a radius is required, forgetting to square a radius in a disk or washer model, ignoring a velocity sign change when finding distance, and using the average-value formula without dividing by interval length.

Interactive Tasks

Quiz: Test Your Knowledge

What does a definite integral represent when a graph crosses the horizontal axis? (Net signed accumulation) (!Only positive geometric area) (!The maximum function value) (!The slope at the midpoint)

For vertical slices, how is the area between two curves found when one curve stays above the other? (Upper function minus lower function) (!Lower function minus upper function) (!Sum of both functions) (!Product of both functions)

What must you usually do to find total geometric area when a function changes sign? (Split at zeros or use absolute value) (!Differentiate the function twice) (!Replace both bounds with zero) (!Multiply the integral by negative one)

Which quantity is integrated to find a volume from known cross sections? (Cross sectional area) (!Boundary slope) (!Average radius) (!Perimeter only)

What distinguishes the washer method from the disk method? (An inner circular area is subtracted) (!The radius is never squared) (!No definite integral is needed) (!Only horizontal axes are allowed)

How is the average value of a continuous function found on an interval? (Integral divided by interval length) (!Integral multiplied by interval length) (!Largest value minus smallest value) (!Derivative divided by interval length)

What does the integral of velocity over time give? (Displacement) (!Total distance in every case) (!Acceleration) (!Average speed in every case)

What extra operation is needed to obtain total distance from velocity when direction changes? (Integrate the absolute value of velocity) (!Differentiate velocity) (!Square the time interval) (!Use only positive bounds)

In one dimensional motion, what does integrating force over displacement calculate? (Work) (!Power) (!Momentum) (!Acceleration)

For a continuous probability density, what gives the probability of an interval? (The integral of the density over that interval) (!The derivative of the density over that interval) (!The maximum density value) (!The interval length alone)

Memory Game

Definite integral	Net accumulation over an interval
Riemann sum	Approximation built from many small contributions
Average value	Integral divided by interval length
Cross section	Slice whose area can be accumulated into volume
Washer method	Outer circular area minus inner circular area
Arc length	Distance measured along a curve
Probability density	Nonnegative function whose interval area gives probability

Drag and Drop

Match the correct terms.	Topic
Top minus bottom	Area between curves with vertical slices
Area times thickness	Volume from cross sections
Velocity times time	Displacement accumulation
Force times displacement	Work accumulation
Integral divided by width	Average value

Crossword Puzzle

Integral	What accumulation tool combines infinitely many small contributions?
Accumulation	What broad idea unifies area, volume, displacement, work, and probability?
Displacement	What net change in position is obtained by integrating velocity?
Washer	Which method subtracts an inner disk from an outer disk?
Arclength	What term names the distance measured along a curve?
Probability	What quantity is obtained by integrating a density over an interval?

LearningApps

Cloze Text

A definite integral measures

over an interval. When a graph lies below the horizontal axis, its signed contribution is

. The Fundamental Theorem of Calculus evaluates many definite integrals using an

. Area between two curves with vertical slices is found from

. A volume model accumulates

through the solid. The average value of a function equals its integral divided by the

. Integrating velocity gives

rather than total distance when direction changes. Integrating a variable force over displacement gives

. The area under a probability density over an interval gives

.

Open-Ended Tasks

Easy

Area sketch: Draw two intersecting functions, shade the enclosed region, identify the intersection points, and write the area integral without evaluating it.
Motion story: Create a short motion story whose velocity changes sign, sketch a matching velocity graph, and explain the difference between displacement and total distance.
Average value infographic: Produce a one-page visual explanation showing why average value is the area-equivalent height of a rectangle.
Signed area video: Record a two-minute explanation of how positive and negative regions affect a definite integral, using your own graph.

Standard

Cross-section model: Build a paper or digital model of a solid with square or semicircular cross sections and derive an integral for its volume.
Solid of revolution design: Choose two functions, create a region, rotate it around an axis, and compare disk and washer descriptions where appropriate.
Calculus interview: Interview an engineer, scientist, economist, or teacher about a situation in which accumulated change matters, then translate one example into an integral model.
Variable force experiment: Use a spring scale or a safe simulated data set to record force at several displacements, estimate work numerically, and compare the result with an integral model.

Advanced

Numerical integration project: Collect or locate time-series data for a rate such as flow, power, or velocity, estimate total accumulation with several numerical rules, and analyze the error.
Arc length investigation: Choose a curved path from a map, design, or graph, approximate its length with line segments, then compare the approximation with an arc-length integral.
Probability simulation: Define a valid continuous density on a finite interval, calculate a probability by integration, simulate many random outcomes, and compare experimental and theoretical proportions.
Application synthesis: Create a mini research presentation that connects at least three integral applications, explains their common accumulation structure, and identifies where the models differ.

Learning Assessment

Model selection: Given several real situations, identify which require area, volume, average value, displacement, distance, work, arc length, or probability, and justify every choice from the meaning of a small contribution.
Error diagnosis: Analyze a flawed solution in which signed area is confused with geometric area or inner and outer washer radii are reversed, then correct both the model and the reasoning.
Units and plausibility: Solve an applied integral problem, track units through the integrand and differential, and use a rough estimate to judge whether the final magnitude is reasonable.
Multiple representations: Represent one application with a graph, a verbal explanation, a Riemann-sum description, and a definite integral, and explain how the four representations correspond.
Transfer problem: Design and solve a new problem in which a measured rate must be accumulated over time, including assumptions, bounds, units, calculation, and interpretation.
Method comparison: For one solid of revolution, compare two valid slicing approaches and argue which setup is more efficient while showing that both describe the same volume.

Evidence of Learning

Evidence of learning should show more than correct antiderivatives. Your work should demonstrate the following.

Knowledge: You can explain definite integrals as limits of sums, distinguish signed accumulation from geometric area, and state the main formulas for area, volume, average value, displacement, distance, work, arc length, and probability.

Skills: You can sketch relevant regions, find intersections and sign changes, choose a variable of integration, construct integrands from small contributions, evaluate or approximate definite integrals, and use units to check meaning.

Products: Useful evidence includes annotated graphs, mathematical models, physical or digital solids, short explanatory videos, experimental data tables, simulations, and written solutions that connect formulas to context.

Transfer achievements: You can recognize accumulation in an unfamiliar situation, adapt a known integral model, justify assumptions, compare alternative methods, and communicate why the result is mathematically and contextually reasonable.

OERs on the Topic

The English Wikipedia article on integrals offers a broad reference for the concept, notation, history, interpretations, and applications.

For further open learning, explore the linked topics Fundamental theorem of calculus, Riemann sum, Solid of revolution, Arc length, work in physics, and Probability density function.

Linked Learning Areas

The following topics connect applications of integrals to core ideas in mathematics and neighboring subjects.

Applications of Integrals

aiMOOC Projects

Diesen aiMOOC als Battle spielen: Mathematik, Klasse 11-13, G-Niveau

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English:Applications of Integrals