Name: integration in breathing cycle modeling
Uploaded: 2026-01-12T12:24:28.000Z
Duration: 1 min 30 s
Channel: JoVE

הנשימה היא תהליך מחזורי, כאשר כל מחזור נשימה מלא נמשך כ-5 שניות.המטרה היא למצוא את נפח האוויר שנשאף בריאות בכל רגע במהלך מחזור זה.קצב זרימת האוויר משתנה עם הזמן וניתן לדגם אותו באמצעות פונקציה סינוסואידלית. הוא מראה שזרימת האוויר מגיעה לשיא באמצע השאיפה ומתהפכת במהלך הנשיפה, עם קצב מרבי של כ-0.5 ליטר לשנייה.הנפח האינפיניטסימלי dV במשך זמן קצר dt מצטבר לנפח הכולל שנשאף על ידי אינטגרציה של קצב זרימת האוויר מ-0 ל-t.שיטת ההחלפה מפשטת את האינטגרל על ידי קביעת המשתנה u כ-2πx חלקי 5. הדיפרנציאל dx מבוטא במונחים של du כדי להתאים למשתנה החדש. גבולות האינטגרציה משתנים גם הם כדי להתאים למשתנה החדש, ונעים בין 0 ל-2πt חלקי 5.האינטגרל הופך לצורת סינוס סטנדרטית, שהאינטגרל שלה הוא קוסינוס שלילי, ומוערכת באמצעות גבולות האינטגרציה.זה נותן את נפח האוויר שנשאף בליטרים כפונקציה של הזמן במהלך מחזור הנשימה, ומגיע למקסימום במהלך השאיפה.מודל זה כולל גם כיוון וגם קצב זרימת האוויר, ומספק תצוגה ברורה של נפח הנשימה לאורך זמן.

יישום אינטגרציה: פתרון בעיות

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&#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8;&#x5D6;&#x5E8;&#x5D9;&#x5DE;&#x5EA; &#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8; &#x5D1;&#x5DE;&#x5D4;&#x5DC;&#x5DA; &#x5D4;&#x5E0;&#x5E9;&#x5D9;&#x5DE;&#x5D4; &#x5D0;&#x5D9;&#x5E0;&#x5D4; &#x5E7;&#x5D1;&#x5D5;&#x5E2;&#x5D4;; &#x5D4;&#x5D9;&#x5D0; &#x5DE;&#x5E9;&#x5EA;&#x5E0;&#x5D4; &#x5E2;&#x5DD; &#x5D4;&#x5D6;&#x5DE;&#x5DF; &#x5D5;&#x5DE;&#x5D9;&#x5D5;&#x5E6;&#x5D2;&#x5EA; &#x5D1;&#x5D0;&#x5D5;&#x5E4;&#x5DF; &#x5D4;&#x5DE;&#x5D3;&#x5D5;&#x5D9;&#x5E7; &#x5D1;&#x5D9;&#x5D5;&#x5EA;&#x5E8; &#x5E2;&#x5DC; &#x5D9;&#x5D3;&#x5D9; &#x5E4;&#x5D5;&#x5E0;&#x5E7;&#x5E6;&#x5D9;&#x5D4; &#x5E1;&#x5D9;&#x5E0;&#x5D5;&#x5E1;&#x5D5;&#x5D0;&#x5D9;&#x5D3;&#x5DC;&#x5D9;&#x5EA;. &#x5D1;&#x5DE;&#x5D7;&#x5D6;&#x5D5;&#x5E8; &#x5E0;&#x5E9;&#x5D9;&#x5DE;&#x5D4; &#x5E8;&#x5D2;&#x5D9;&#x5DC;, &#x5E7;&#x5E6;&#x5D1; &#x5D6;&#x5E8;&#x5D9;&#x5DE;&#x5EA; &#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8; &#x5DE;&#x5D2;&#x5D9;&#x5E2; &#x5DC;&#x5DE;&#x5E7;&#x5E1;&#x5D9;&#x5DE;&#x5D5;&#x5DD; &#x5E9;&#x5DC; &#x5DB;&#x5BE;0.5 L/s &#x5D1;&#x5E0;&#x5E7;&#x5D5;&#x5D3;&#x5EA; &#x5D4;&#x5D0;&#x5DE;&#x5E6;&#x5E2; &#x5E9;&#x5DC; &#x5D4;&#x5E9;&#x5D0;&#x5D9;&#x5E4;&#x5D4; &#x5D5;&#x5D9;&#x5D5;&#x5E8;&#x5D3; &#x5D1;&#x5D0;&#x5D5;&#x5E4;&#x5DF; &#x5E1;&#x5D9;&#x5DE;&#x5D8;&#x5E8;&#x5D9; &#x5DC;&#x5DB;&#x5D9;&#x5D5;&#x5D5;&#x5DF; &#x5D0;&#x5E4;&#x5E1; &#x5D1;&#x5EA;&#x5D7;&#x5D9;&#x5DC;&#x5EA; &#x5D5;&#x5D1;&#x5E1;&#x5D5;&#x5E3; &#x5DB;&#x5DC; &#x5E9;&#x5DC;&#x5D1;. &#x5D4;&#x5EA;&#x5E0;&#x5D4;&#x5D2;&#x5D5;&#x5EA; &#x5EA;&#x5E7;&#x5D5;&#x5E4;&#x5EA;&#x5D9;&#x5EA; &#x5D6;&#x5D5; &#x5E0;&#x5D9;&#x5EA;&#x5E0;&#x5EA; &#x5DC;&#x5DE;&#x5D9;&#x5D3;&#x5D5;&#x5DC; &#x5D1;&#x5D0;&#x5DE;&#x5E6;&#x5E2;&#x5D5;&#x5EA; &#x5D4;&#x5E4;&#x5D5;&#x5E0;&#x5E7;&#x5E6;&#x5D9;&#x5D4;:\begin{equation*}\jfrac{dV}{dt} = 0.5 \sin\liparens{\jfrac{2\pi t}{5}}\end{equation*}&#x5DB;&#x5D0;&#x5E9;&#x5E8; dV/dt &#x5D4;&#x5D5;&#x5D0; &#x5E7;&#x5E6;&#x5D1; &#x5D6;&#x5E8;&#x5D9;&#x5DE;&#x5EA; &#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8; &#x5D1;&#x5DC;&#x5D9;&#x5D8;&#x5E8;&#x5D9;&#x5DD; &#x5DC;&#x5E9;&#x5E0;&#x5D9;&#x5D9;&#x5D4;, &#x5D5;&#x5BE;t &#x5D4;&#x5D5;&#x5D0; &#x5D4;&#x5D6;&#x5DE;&#x5DF; &#x5D1;&#x5E9;&#x5E0;&#x5D9;&#x5D5;&#x5EA;.&#x5D0;&#x5D9;&#x5E0;&#x5D8;&#x5D2;&#x5E8;&#x5E6;&#x5D9;&#x5D4; &#x5D1;&#x5D0;&#x5DE;&#x5E6;&#x5E2;&#x5D5;&#x5EA; &#x5D4;&#x5E6;&#x5D1;&#x5D4;&#x5DB;&#x5D3;&#x5D9; &#x5DC;&#x5E7;&#x5D1;&#x5D5;&#x5E2; &#x5D0;&#x5EA; &#x5E0;&#x5E4;&#x5D7; &#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8; &#x5D4;&#x5E0;&#x5E9;&#x5D0;&#x5E3; V(t) &#x5D1;&#x5DB;&#x5DC; &#x5D6;&#x5DE;&#x5DF; t, &#x5D9;&#x5E9; &#x5DC;&#x5D1;&#x5E6;&#x5E2; &#x5D0;&#x5D9;&#x5E0;&#x5D8;&#x5D2;&#x5E8;&#x5E6;&#x5D9;&#x5D4; &#x5E9;&#x5DC; &#x5E4;&#x5D5;&#x5E0;&#x5E7;&#x5E6;&#x5D9;&#x5D9;&#x5EA; &#x5D6;&#x5E8;&#x5D9;&#x5DE;&#x5EA; &#x5D4;&#x5D0;&#x5D5;&#x5D5;&#x5D9;&#x5E8;:\begin{equation*}V(t) = \int_0^t 0.5 \sin\liparens{\jfrac{2\pi x}{5}}dx\end{equation*}&#x5D1;&#x5D0;&#x5DE;&#x5E6;&#x5E2;&#x5D5;&#x5EA; &#x5D4;&#x5D4;&#x5E6;&#x5D1;&#x5D4; 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&#x5EA;&#x5D4;&#x5DC;&#x5D9;&#x5DA; &#x5D4;&#x5E0;&#x5E9;&#x5D9;&#x5DE;&#x5D4; &#x5DB;&#x5D5;&#x5DC;&#x5DC; &#x5DB;&#x5E0;&#x5D9;&#x5E1;&#x5D4; ...

application-of-integration-problem-solving

Education

JoVE Core

Calculus

חשבון דיפרנציאלי ואינטגרלי

Unit

יחידה

Integrals

אינטגרלים

KEY TERMS AND CONCEPTS

מונחי מפתח ומושגים

Application of Integration: Problem Solving

integration in breathing cycle modeling

Integration can be used to model the volume of air inhaled during breathing. The respiratory cycle is the repeating intake and expulsion of air, and it usually lasts about five seconds. By treating airflow as a function of time, students can see how the amount of air in the lungs changes during a breath.

Airflow during respiration is not constant. It changes in a smooth, wave-like pattern, so a sinusoidal function is a good model. In a normal breathing cycle, the airflow rate reaches a maximum of about 0.5 liters per second at the midpoint of inhalation. It then falls symmetrically to zero at the start and end of each phase.

To find the inhaled volume V(t), the airflow rate is integrated over time. The model uses dV/dt in liters per second and time t in seconds. With the substitution u = 2πx/5, the integral becomes easier to solve. The result gives a function for the volume of inhaled air over time.

This integral describes how much air enters the lungs as breathing continues. It also captures both the size and direction of airflow. The volume curve peaks at mid-inhalation, which matches the normal pattern of lung movement during a typical breath cycle.

Integration in Breathing Cycle Modeling

area problem

Approximating Area Under a Curve

Approximating area under a curve is a key way to measure regions with curved boundaries. Straight-sided shapes are easy to handle with formulas for rectangles, triangles, and polygons. Curved regions need a different method because standard geometry does not work well on them. The area problem solves this by breaking the region into smaller, simpler shapes. A common approach uses rectangles to estimate the space under a function. Each rectangle gets its height from the function at a chosen...

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Estimating Distance from Changing Velocity

Estimating distance from changing velocity uses small time intervals and velocity values. When an object speeds up or slows down, its total displacement can be found by adding many short pieces of motion. This method gives an approximation first, and integral calculus gives the exact result later. A runner who accelerates during the first three seconds of a race shows how the idea works. Velocity can be measured every half-second and multiplied by the 0.5-second time step. Using left-endpoint...

definite integral

Riemann Sums and Area Estimates

Riemann sums and area estimates are a key part of calculus for finding the area under a curve. When a real-valued function is defined on a closed interval, the area under its graph can be approximated if an exact calculation is not easy to do. The interval is split into equal subintervals. Each subinterval becomes a rectangle. The rectangle’s width is the length of the subinterval, and its height comes from the function value at one chosen point inside that part of the interval. Multiplying...

sums of power

Gauss Sums for Riemann Rectangles

Riemann sums use rectangles to estimate the area under a curve in definite integration. The interval is split into subintervals, and the rectangle areas are added together. When those rectangle heights follow a regular pattern, sum formulas can make the calculation faster and more exact. This page focuses on sums that appear in arithmetic and polynomial sequences. The sum of consecutive integers, squares, and cubes is especially useful when the partition is uniform or when the data has a...

midpoint rule

Estimating Area with the Midpoint Rule

The Midpoint Rule estimates the area under a curve by using rectangles. It is a numerical method that helps when an exact area is hard to find or not practical to calculate. By using rectangle heights from midpoints, it gives a useful approximation over a chosen interval. To use the Midpoint Rule, the interval is split into equal subintervals. The midpoint of each subinterval is found by averaging its two endpoints. The function value at that midpoint gives the height of a rectangle over that...

properties of definite integral i

Definite Integrals for Car Motion

Definite integrals help describe a car’s motion over time by linking velocity to displacement. In a velocity-time graph, the definite integral shows how velocity builds up over a chosen time interval. The area under the curve gives the total displacement, so the graph becomes a clear picture of how far the car moves. One key idea is linearity of definite integrals. This property lets you add or subtract velocity functions while keeping their accumulated effect. If a car’s motion is made of...

properties of definite integral ii

Definite Integrals in Motion Analysis

Definite integrals can be used to find displacement from a velocity-time graph. A velocity function v(t) describes how motion changes over time. The definite integral of v(t) from a to b gives the net displacement between those times. It also represents the signed area under the velocity curve between the two points. Two useful properties make definite integrals easier to work with in motion problems. The additivity property says that if a continuous function is integrated over adjacent...

properties of definite integral iii

Definite Integral Rules for Motion

The definite integral helps describe motion by showing how far an object travels over time. In this lesson, the key ideas are the Positivity Property and the Comparison Property of definite integrals. Both ideas connect integration to velocity and displacement in a way that is easy to interpret in real situations. The Positivity Property says that if an object’s velocity is nonnegative, then its total displacement over the time interval is also nonnegative. Nonnegative velocity means the...

fundamental theorem of calculus i

Definite Integrals: A Step-by-Step Strategy

Definite integrals are solved best with a clear, step-by-step strategy. The first step is to identify what the problem is asking for. The quantity may be accumulation, area, force, or probability. It is also important to note the integration limits and any constraints that affect the setup. Before evaluating the integral, look for properties that can simplify the work. If the expression has multiple parts, break it into smaller integrals. Known values from earlier work may help too. Symmetry...

fundamental theorem of calculus i: problem solving

fundamental-theorem-of-calculus-i-problem-solving

Accumulation and Rates in the Fundamental Theorem

The Fundamental Theorem of Calculus connects accumulation and rate of change. In this lesson, the idea is shown with a water tank that is filled by a pump. The pump rate changes over time, so the total water collected must be found by adding up the changing inflow. In water management, this kind of calculation matters for pressure control, operation timing, and system safety. The flow rate is given by a time-dependent function. To find how much water has entered the tank from the start to a...

fundamental theorem of calculus ii

Definite Integrals with Antiderivatives

The Fundamental Theorem of Calculus, Part 2, connects definite integrals with antiderivatives. It shows that the area under a continuous curve can be found without adding infinitely many tiny rectangles one by one. Instead, the integral can be evaluated directly from an antiderivative. If a function f(x) is continuous on a closed interval [a, b], then its definite integral over that interval can be computed using any antiderivative F(x). The result is written as F(b) - F(a). This works because...

application of integration: problem solving

Integration can be used to model the volume of air inhaled during breathing. The respiratory cycle is the repeating intake and expulsion of air, and it usually lasts about five seconds. By treating airflow as a function of time, students can see how the amount of air in the lungs changes during a breath. Airflow during respiration is not constant. It changes in a smooth, wave-like pattern, so a sinusoidal function is a good model. In a normal breathing cycle, the airflow rate reaches a maximum...

indefinite integrals

Finding Water Volume with Integration

Integration can be used to find the total volume of water in a tank when the inflow rate changes over time. In this example, the pump starts at 5 L/min. The inflow rate then increases by 2 L/min for each additional minute. That changing rate can be written as a linear function. To measure how much water has been added, the inflow rate function is integrated over time. The total water volume V(t) comes from this integral. When the power rule is used, the result includes both a quadratic term...

net change theorem

Calculus for Reservoir Overflow Prediction

The Net Change Theorem links a function’s rate of change to its total accumulated change over an interval. In calculus, this means the definite integral of a derivative on [a,b] gives the net change in the original function. The theorem is useful in physics, economics, and engineering because it turns a rate into a total amount. One clear application is a reservoir that receives water from two inflow sources at the same time. The reservoir starts with 260 million cubic meters of water. Each...

substitution rule applied to indefinite integrals

substitution-rule-applied-to-indefinite-integrals

Work Done on a Nonlinear Spring

Work done on a nonlinear spring can be found by using integration and the substitution rule. A linear spring follows Hooke’s law, so its restoring force increases in direct proportion to displacement. That makes the force-displacement graph a straight line and the work calculation straightforward. A nonlinear spring does not obey Hooke’s law. Its restoring force changes with position in a non-proportional way, so the force-displacement graph is curved instead of linear. In the example shown,...

substitution rule applied to definite integrals

substitution-rule-applied-to-definite-integrals

Using Substitution to Evaluate Definite Integrals

Substitution makes definite integrals easier to evaluate when the integrand is a composite function. It works by reversing the chain rule from differentiation. When the expression includes an inner function and its derivative, substitution can simplify the setup quickly. The method starts by choosing the inner function and naming it with a new variable. Then its differential is found and used to replace the original variable terms. The limits of integration are also changed by plugging the...

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growth-models-with-integration-problem-solving

Finding Fish Biomass with Integrals

Fish biomass can be modeled with integration when the growth rate is known over time. In ecology, biomass means the total weight of a fish population in a body of water. If the growth rate of that biomass is given as a function of time, calculus can be used to find the total biomass at a later date. The growth rate is written as G(t), where t is time in years and G(t) is the rate of change of biomass in kilograms per year. The biomass itself is written as B(t). Because the growth rate shows...

key terms and concepts

See how integration models inhaled air in the breathing cycle using a sinusoidal airflow function, substitution, and a volume curve for lung ventilation.

Integration in Breathing Cycle Modeling | JoVE K12

The JoVE Core Calculus collection provides a comprehensive visual guide to the entire calculus sequence, seamlessly connecting mathematical theory to real-world applications in physics and economics. Through dynamic animated lessons, the course transitions smoothly from foundational single-variable concepts to advanced multivariable topics. Students will explore everything from standard differentiation and integration to spatial vectors, multi-dimensional derivatives, and complex vector fields. By breaking down abstract coordinate systems, infinite series, and spatial motion, this structured approach builds the robust analytical skills and advanced mastery essential for STEM success.

calculus

Integration in Breathing Cycle Modeling

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