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Chapter 5: Problem 10

Explain why \(\int_{a}^{b} f^{\prime}(x) d x=f(b)-f(a)\)

Short Answer

Expert verified
Question: Explain why the integral of a function's derivative from a to b is equal to the difference of the function's values at these points. Answer: The integral of a function's derivative from a to b is equal to the difference of the function's values at these points because of the Fundamental Theorem of Calculus. When we define a function F(x) as the integral of the derivative of another function f(x) and differentiate F(x), we get back the original derivative, f'(x). Integrating F'(x) from a to b yields F(b) - F(a), and by substituting f'(x) for F'(x) and using the definition of F(x), we can see that the integral of f'(x) from a to b equals f(b) - f(a).

Step by step solution

01

Define the function F(x)

Let F(x) be the function defined as the integral of the derivative of f(x) with respect to x, from a to x, i.e.: \(F(x) = \int_{a}^{x} f^{\prime}(t) dt\)
02

Differentiate F(x) with respect to x

By the Fundamental Theorem of Calculus part 1, the derivative of F(x) with respect to x is the original function's derivative, f'(x). That is, \(F'(x) = f^{\prime}(x)\).
03

Integrate F'(x) with respect to x from a to b

Now, we will integrate F'(x) with respect to x from a to b, using the Fundamental Theorem of Calculus part 2: \(\int_{a}^{b} F^{\prime}(x) dx = F(b) - F(a)\)
04

Substitute f'(x) for F'(x) and use the definition of F(x) to show the result

Since \(F'(x) = f^{\prime}(x)\), we can substitute f'(x) in place of F'(x) in the above equation: \(\int_{a}^{b} f^{\prime}(x) dx = F(b) - F(a)\) Since we defined \(F(x) = \int_{a}^{x} f^{\prime}(t) dt\), we can replace F(a) and F(b) with their respective definitions: \(\int_{a}^{b} f^{\prime}(x) dx = \left(\int_{a}^{b} f^{\prime}(t) dt\right) - \left(\int_{a}^{a} f^{\prime}(t) dt\right)\) Now, we can notice that integrating from a to a in the second term would result in no accumulation of area and therefore give a value of 0: \(\int_{a}^{b} f^{\prime}(x) dx = \left(\int_{a}^{b} f^{\prime}(t) dt\right) - 0 = \int_{a}^{b} f^{\prime}(t) dt\) Thus, we can see that \(\int_{a}^{b} f^{\prime}(x) dx = f(b) - f(a)\), demonstrating the validity of the Fundamental Theorem of Calculus.

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Key Concepts

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

Integration
Integration is a core concept in calculus that is used to find the total accumulation of a quantity, such as area under a curve. Fundamentally, integration is the inverse operation to differentiation. This means that if you differentiate a function and then integrate that derivative, you should expect to get back to your original function.

When you integrate a function, you're essentially adding up an infinite number of infinitesimally small quantities. These quantities could represent areas, volumes, or any other measure depending on the context of the problem. Integration allows us to solve a wide range of problems in physics, engineering, economics, and beyond, giving us the power to calculate things like accumulated distance based on velocity or the total growth from a rate of change.
Derivatives
Derivatives represent the rate at which a quantity changes. In a graphical sense, the derivative of a function at any point gives us the slope of the tangent line to the function at that point. This concept is fundamental to understanding how things change over time or space.

For example, if we have a function that describes the position of a car over time, taking the derivative of that function would give us the velocity of the car - how quickly its position is changing. The process of finding a derivative is called differentiation. In the context of the Fundamental Theorem of Calculus, the existence of a derivative tells us that the underlying function is continuous and thus integrable over the interval in question.
Antiderivatives
An antiderivative is essentially the reverse of a derivative, representing a function whose derivative is the given function. Finding an antiderivative is a process also known as integration or indefinite integration. Unlike a derivative, which is unique, antiderivatives of a function are not unique; they differ by a constant, since differentiating a constant gives zero.

This concept of antiderivatives is crucial when working with the Fundamental Theorem of Calculus, as it allows us to understand the relationship between differentiation and integration. In the exercise at hand, an antiderivative helps us to find the original function given its derivative, which is then used to evaluate the definite integral.
Definite Integrals
Definite integrals provide a way to calculate the exact accumulation of quantities, such as area under a curve, between specific bounds. Unlike indefinite integrals, which are families of functions defined up to an added constant, a definite integral calculates a numerical value that represents this total accumulation over an interval \( [a, b] \).

This specific concept is pivotal in applying the Fundamental Theorem of Calculus. It allows us to evaluate the integral of a function's derivative over an interval and directly relate it to the values of the original function at the boundary points through \( f(b) - f(a) \). In practical terms, this means that if you know how a quantity is changing at every moment (the derivative), you can calculate how much it has changed over a period of time (the definite integral).

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Most popular questions from this chapter

Substitutions Suppose that \(p\) is a nonzero real number and \(f\) is an odd integrable function with \(\int_{0}^{1} f(x) d x=\pi .\) Evaluate each integral. a. \(\int_{0}^{\pi /(2 p)} \cos p x f(\sin p x) d x\) b. \(\int_{-\pi / 2}^{\pi / 2} \cos x f(\sin x) d x\)

Use a change of variables to find the following indefinite integrals. Check your work by differentiating. $$\int \frac{(\sqrt{x}+1)^{4}}{2 \sqrt{x}} d x$$

Substitution: shift Perhaps the simplest change of variables is the shift or translation given by \(u=x+c,\) where \(c\) is a real number. a. Prove that shifting a function does not change the net area under the curve, in the sense that $$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(u) d u$$ b. Draw a picture to illustrate this change of variables in the case that \(f(x)=\sin x, a=0, b=\pi,\) and \(c=\pi / 2\)

Find the following integrals. $$\int(z+1) \sqrt{3 z+2} d z$$

If necessary, use two or more substitutions to find the following integrals. \(\int \frac{d x}{\sqrt{1+\sqrt{1+x}}}(\text {Hint: Begin with } u=\sqrt{1+x} .)\)
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