Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How is integration related to derivation?

Short Answer

Expert verified
Question: Explain the relationship between integration and differentiation, and provide an example. Answer: Integration and differentiation are inverse operations, meaning that one process essentially undoes the other. They are two fundamental concepts in calculus; differentiation is the process of finding the derivative or rate of change of a function, and integration is the process of finding the antiderivative or integral of a function. If you have the derivative of a function (f'(x)), and you integrate it, you will get back the original function (f(x)) plus a constant (C). For example, if f(x) = 2x^2, its derivative, f'(_x), equals 4x. Integrating f'(x) gives back the original function, ∫4x dx = 2x^2 + C.

Step by step solution

01

Define differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change or slope of a function at a given point. In mathematical terms, if a function is given by f(x), then its derivative is denoted by f'(x) or df/dx, and it measures the change in y (output) with respect to x (input).
02

Define integration

Integration is the opposite process of differentiation. It is the process of finding the antiderivative or the integral of a given function. The integral represents the net area under the curve of a function. In mathematical terms, suppose F(x) is the antiderivative of f(x), then the integral of f(x) is written as ∫f(x)dx = F(x) + C, where C is the constant of integration.
03

Explain the relationship between differentiation and integration

Integration and differentiation are inverse operations, meaning that one process essentially undoes the other. If you have the derivative of a function (f'(x)), and you integrate it, you will get back the original function (f(x)) plus a constant (C).
04

Provide an example

Let's consider a simple example: - Let f(x) = 2x^2 - Find the derivative, f'(x), by applying the power rule: f'(x) = 4x (using the power rule where the derivative of ax^n = nax^(n-1)) - Now, integrate f'(x) with respect to x: ∫4x dx = 2x^2 + C (using the power rule of integration, ∫x^n dx = (x^(n+1))/(n+1) + C) As shown in this example, after differentiating the given function (f(x)), and then integrating its derivative (f'(x)), we obtained the original function (2x^2) plus a constant (C). This demonstrates the inverse relationship between differentiation and integration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus, pivotal in understanding various physical phenomena. It enables us to find the derivative of a function, which essentially represents the rate at which a function's value changes at any given point. Let's say you have a function \( f(x) \) that represents a curve. Differentiation allows you to find \( f'(x) \), the function's derivative, which tells us how steep or flat the curve is at a specific point. This is particularly useful in physics to calculate speed from distance functions, or in economics to understand marginal cost changes.

Derivatives can be visualized as the slope of the tangent line to the curve at any point \( x \). This gives an instantaneous rate of change, which is crucial in fields such as engineering and natural sciences. By applying rules like the power rule, product rule, or chain rule, you can find the derivative for more complex functions. Differentiation transforms a function to reveal subtle information about its behavior and trends.
Integration
Integration is often referred to as the process that is opposite to differentiation. While differentiation breaks down functions to find rates of change, integration combines small quantities to determine whole quantities like area or volume. In mathematical terms, integration is about finding the antiderivative of a function. If you’re given a derivative, \( f'(x) \), then by integrating it, you can find its original function \( f(x) \).

The symbol for integration is \( \int \), which stands for integrating a function with respect to a variable. The result is often expressed together with a constant \( C \), which represents any constant that might have been part of the original function but disappeared upon differentiation. This constant is crucial for representing the family of all antiderivatives.

Integration allows us to calculate the area under the curve of \( f(x) \) in a coordinate system, which is practical for real-world problems such as finding the total distance traveled given a velocity function. Techniques like substitution and integration by parts help solve more complex integrals.
Inverse Operations
Differentiation and integration are known as inverse operations in calculus. This means that each one essentially reverses the process of the other. If you have a function \( f(x) \), and you differentiate it to get \( f'(x) \), then integrating \( f'(x) \) will bring you back to \( f(x) \) along with a constant \( C \).

This inverse relationship is similar to how addition and subtraction reverse each other. When a function's derivative is known, integrating it acts as a process of reconstructing the original function. In mathematical equations, if you start with \( f(x) \), differentiate into \( f'(x) \), and then integrate, you'll need to remember that the constant \( C \) keeps track of any potential vertical shifts in the function.

Understanding this relationship is key in calculus problems that require finding a function from its rate of change or vice versa. This principle helps solve a wide range of practical problems, including motion dynamics in physics and calculating growth rates in biology and economics.
Antiderivative
An antiderivative is a function that reverses the process of differentiation. It provides insight into pinpointing the original function from which a given derivative was derived. If you know \( F(x) \) is the antiderivative of \( f'(x) \), differentiating \( F(x) \) will yield \( f(x) \). This concept is pivotal when dealing with integration problems.

The notation \( \int f(x) \, dx = F(x) + C \) means that \( F(x) \) is an antiderivative of \( f(x) \). Here, \( C \) represents the constant of integration described earlier. Antiderivatives are not unique but belong to a family of functions that differ by a constant \( C \).

In practical terms, finding an antiderivative allows us to construct a broader picture of the function's behavior, such as calculating cumulative quantities from rates, seen in contexts like energy expenditure over time or cost accumulation. Recognizing or deriving correct antiderivatives is essential in evaluating integrals accurately in different domains, from physics to finance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Is the thermal conductivity of a medium, in general, constant or does it vary with temperature?

The outer surface of an engine is situated in a place where oil leakage can occur. Some oils have autoignition temperatures of approximately above \(250^{\circ} \mathrm{C}\). When oil comes in contact with a hot engine surface that has a higher temperature than its autoignition temperature, the oil can ignite spontaneously. Treating the engine housing as a plane wall, the inner surface \((x=0)\) is subjected to \(6 \mathrm{~kW} / \mathrm{m}^{2}\) of heat. The engine housing \((k=13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a thickness of \(1 \mathrm{~cm}\), and the outer surface \((x=L)\) is exposed to an environment where the ambient air is \(35^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event the leaked oil comes in contact with the hot engine surface, the temperature of the engine surface should be kept below \(200^{\circ} \mathrm{C}\). Determine the variation of temperature in the engine housing and the temperatures of the inner and outer surfaces. Is the outer surface temperature of the engine below the safe temperature?

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{1}=80^{\circ} \mathrm{C}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperature of the right surface of the wall at \(x=L\).

Consider the uniform heating of a plate in an environment at a constant temperature. Is it possible for part of the heat generated in the left half of the plate to leave the plate through the right surface? Explain.

Consider a spherical reactor of \(5-\mathrm{cm}\) diameter operating at steady condition has a temperature variation that can be expressed in the form of \(T(r)=a-b r^{2}\), where \(a=850^{\circ} \mathrm{C}\) and \(b=5 \times 10^{5} \mathrm{~K} / \mathrm{m}^{2}\). The reactor is made of material with \(c=\) \(200 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=9000 \mathrm{~kg} / \mathrm{m}^{3}\). If the heat generation of reactor is suddenly set to \(9 \mathrm{MW} / \mathrm{m}^{3}\), determine the time rate of temperature change in the reactor. Is the heat generation of reactor suddenly increased or decreased to \(9 \mathrm{MW} / \mathrm{m}^{3}\) from its steady operating condition?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free