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Chapter 4: Problem 53

State the Fundamental Theorem of Calculus.

Short Answer

Expert verified
The Fundamental Theorem of Calculus consists of two parts: 1) If a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). 2) If a function f is continuous over an interval I containing a, then for every x in I, the function F defined by \( F(x) = \int_{a}^{x} f(t) \, dt \) is continuous on I and differentiable on the interior of I, and F'(x) = f(x) for all x in I.

Step by step solution

01

Formulate Part 1

The first part of the Fundamental Theorem of Calculus, often called the First Fundamental Theorem of Calculus, links differentiation and integration in a precise way. It states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). In other words, the definite integral of a function can be calculated using one of its antiderivatives.
02

Formulate Part 2

The second part of the Fundamental Theorem of Calculus, often referred to as the Second Fundamental Theorem of Calculus, gives an integral as a function and describes the derivative of that function. It states that if a function f is continuous over an interval I containing a, then for every x in I, the function F defined by \( F(x) = \int_{a}^{x} f(t) \, dt \) is continuous on I and differentiable on the interior of I, and F'(x) = f(x) for all x in I. This means that the derivative of the integral of a function is just the original function.

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Key Concepts

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

Differentiation
Differentiation is one of the two main operations in calculus. It refers to finding the derivative of a function, which represents the rate of change of the function's value with respect to a variable. The derivative gives us a precise value for how a small change in one variable affects a change in another variable.
  • For example, if you have a function describing the position of a car over time, the derivative gives you the speed of the car at any given moment.
  • In mathematical terms, if you have a function \( f(x) \), the derivative is noted as \( f'(x) \).
The process of differentiation breaks down a complex relationship into simpler rates of change. Understanding differentiation is crucial because it allows us to analyze and predict how a system behaves under different conditions.
Integration
Integration is the inverse process of differentiation, and together they form the core of calculus. Integration involves finding an integral, which represents the accumulation of quantities. It can be thought of as finding the area under a curve.
  • There are two types of integrals: indefinite and definite.
  • Indefinite integrals give us a family of functions and are often expressed with a constant \( C \).
  • Definite integrals provide a specific number that represents the area under a curve from one point to another.
Integration helps in solving problems related to the total amount or cumulative effect, such as calculating distances, areas, and even probabilities in certain cases.
Definite Integral
A definite integral is a type of integral that computes the net area under a function over a specific interval. Unlike indefinite integrals, which yield functions, definite integrals provide numerical values.
  • The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \).
  • The limits of integration, \( a \) and \( b \), represent the bounds of the interval over which you want to find the area.
  • Definite integrals are used for calculating real-world quantities like total distance, area, and volume.
Using the Fundamental Theorem of Calculus, you can calculate definite integrals by evaluating an antiderivative at the upper and lower bounds of the interval.
Antiderivative
Antiderivatives are functions that "reverse" differentiation. Given a function \( f(x) \), its antiderivative \( F(x) \) is a function such that \( F'(x) = f(x) \). Every function has many antiderivatives differing by a constant \( C \).
  • An antiderivative provides a way to backtrack through the process of differentiation, helping to recover a function that gives \( f(x) \) upon differentiation.
  • In practical terms, an antiderivative gives us the "original" function before differentiation occurred.
  • Antiderivatives are crucial for solving integrals. Knowing an antiderivative enables you to compute definite and indefinite integrals.
Effectively working with antiderivatives is central to applying the Fundamental Theorem of Calculus in real-world scenarios.

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Most popular questions from this chapter

Consider a particle moving along the \(x\) -axis where \(x(t)\) is the position of the particle at time \(t, x^{\prime}(t)\) is its velocity, and \(\int_{a}^{b}\left|x^{\prime}(t)\right| d t\) is the distance the particle travels in the interval of time. The position function is given by \(x(t)=t^{3}-6 t^{2}+9 t-2\), \(0 \leq t \leq 5 .\) Find the total distance the particle travels in 5 units of time.

A differential equation, a point, and a slope field are given. A slope field (or direction field) consists of line segments with slopes given by the differential equation. These line segments give a visual perspective of the slopes of the solutions of the differential equation. (a) Sketch two approximate solutions of the differential equation on the slope field, one of which passes through the indicated point. (To print an enlarged copy of the graph, select the MathGraph button.) (b) Use integration to find the particular solution of the differential equation and use a graphing utility to graph the solution. Compare the result with the sketches in part (a). $$ \frac{d y}{d x}=\cos x,(0,4) $$

Use the Second Fundamental Theorem of Calculus to find \(\boldsymbol{F}^{\prime}(\boldsymbol{x})\). $$ F(x)=\int_{0}^{x} t \cos t d t $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(F^{\prime}(x)=G^{\prime}(x)\) on the interval \([a, b]\), then \(F(b)-F(a)=G(b)-G(a)\).

Complete the table. Original Integral $$\int \frac{1}{2 x^{3}} d x$$
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