# 5.4 Integration formulas and the net change theorem

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• Apply the basic integration formulas.
• Explain the significance of the net change theorem.
• Use the net change theorem to solve applied problems.
• Apply the integrals of odd and even functions.

In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.

## Basic integration formulas

Recall the integration formulas given in [link] and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.

## Integrating a function using the power rule

Use the power rule to integrate the function ${\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt.$

The first step is to rewrite the function and simplify it so we can apply the power rule:

$\begin{array}{cc}{\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt\hfill & ={\int }_{1}^{4}{t}^{1\text{/}2}\left(1+t\right)dt\hfill \\ \\ & ={\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt.\hfill \end{array}$

Now apply the power rule:

$\begin{array}{cc}{\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt\hfill & ={\left(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2}\right)|}_{1}^{4}\hfill \\ & =\left[\frac{2}{3}{\left(4\right)}^{3\text{/}2}+\frac{2}{5}{\left(4\right)}^{5\text{/}2}\right]-\left[\frac{2}{3}{\left(1\right)}^{3\text{/}2}+\frac{2}{5}{\left(1\right)}^{5\text{/}2}\right]\hfill \\ & =\frac{256}{15}.\hfill \end{array}$

Find the definite integral of $f\left(x\right)={x}^{2}-3x$ over the interval $\left[1,3\right].$

$-\frac{10}{3}$

## The net change theorem

The net change theorem    considers the integral of a rate of change . It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

## Net change theorem

The new value of a changing quantity equals the initial value plus the integral of the rate of change:

$\begin{array}{}\\ \\ F\left(b\right)=F\left(a\right)+{\int }_{a}^{b}F\text{'}\left(x\right)dx\hfill \\ \hfill \text{or}\hfill \\ {\int }_{a}^{b}F\text{'}\left(x\right)dx=F\left(b\right)-F\left(a\right).\hfill \end{array}$

Subtracting $F\left(a\right)$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement .

We looked at a simple example of this in The Definite Integral . Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in [link] .

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