The Definite Integral: Learn It 1

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The Definite Integral: Learn It 1

Recognize the parts of an integral and when it can be used
Explain how definite integrals relate to the net area under a curve and use geometry to evaluate them
Determine the average value of a function

Defining and Evaluating Definite Integrals

In the preceding section we defined the area under a curve in terms of Riemann sums:

[latex]A=\underset{n\to \infty }{\lim} \displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x[/latex].

However, this definition came with restrictions. We required [latex]f(x)[/latex] to be continuous and nonnegative. Unfortunately, real-world problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.

Definition and Notation

The definite integral generalizes the concept of the area under a curve. We relax the requirement that [latex]f(x)[/latex] be continuous and nonnegative and define the definite integral as follows.

definite integral

If [latex]f(x)[/latex] is a function defined on an interval [latex][a,b][/latex], the definite integral of [latex]f[/latex] from [latex]a[/latex] to [latex]b[/latex] is given by

[latex]\displaystyle\int_a^b f(x) dx=\underset{n\to \infty }{\lim} \displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x[/latex],

provided the limit exists.

If this limit exists, the function [latex]f(x)[/latex] is said to be integrable on [latex][a,b][/latex], or is an integrable function.

The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the [latex]a[/latex] and [latex]b[/latex] above and below) to represent an antiderivative.

Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same.

A definite integral is a number. An indefinite integral is a family of functions.

Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.

Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the codiscoverer of calculus, along with Isaac Newton.

The integration symbol [latex]\displaystyle\int[/latex] is an elongated S, suggesting sigma or summation. On a definite integral, the bounds [latex]a[/latex] and [latex]b[/latex] are the limits of integration, specifying the interval [latex][a,b][/latex]; specifically, [latex]a[/latex] is the lower limit and [latex]b[/latex] is the upper limit.

To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as [latex]n\to \infty[/latex]. Second, the boundaries of the region are called the limits of integration.

We call the function [latex]f(x)[/latex] the integrand, and the [latex]dx[/latex] indicates that [latex]f(x)[/latex] is a function with respect to [latex]x[/latex], called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:

[latex]\displaystyle\int_a^b f(x) dx=\displaystyle\int_a^b f(t) dt=\displaystyle\int_a^b f(u) du[/latex]

Previously, we discussed the fact that if [latex]f(x)[/latex] is continuous on [latex][a,b][/latex], then the limit

[latex]\underset{n\to \infty }{\lim} \displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x[/latex]

exists and is unique.

integrability of continuous functions

If [latex]f(x)[/latex] is continuous on [latex][a,b][/latex], then [latex]f[/latex] is integrable on [latex][a,b][/latex].

Functions that are not continuous on [latex][a,b][/latex] may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.

It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary.

Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero.

This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

Use the definition of the definite integral to evaluate [latex]\displaystyle\int_0^2 x^2 dx[/latex]. Use a right-endpoint approximation to generate the Riemann sum.

We first want to set up a Riemann sum.

Based on the limits of integration, we have [latex]a=0[/latex] and [latex]b=2[/latex]. For [latex]i=0,1,2, \cdots ,n[/latex], let [latex]P=\{x_i\}[/latex] be a regular partition of [latex][0,2][/latex]. Then,

[latex]\Delta x=\frac{b-a}{n}=\frac{2}{n}[/latex].

Since we are using a right-endpoint approximation to generate Riemann sums, for each [latex]i[/latex], we need to calculate the function value at the right endpoint of the interval [latex][x_{i-1},x_i][/latex].

The right endpoint of the interval is [latex]x_i[/latex], and since [latex]P[/latex] is a regular partition,

[latex]x_i=x_0+i \Delta x=0+i(\frac{2}{n})=\frac{2i}{n}[/latex].

Thus, the function value at the right endpoint of the interval is,

[latex]f(x_i)=x_i^2=(\frac{2i}{n})^2=\frac{4i^2}{n^2}[/latex].

Then the Riemann sum takes the form,

[latex]\displaystyle\sum_{i=1}^{n} f(x_i)\Delta x=\displaystyle\sum_{i=1}^{n} (\frac{4i^2}{n^2})\frac{2}{n}=\displaystyle\sum_{i=1}^{n} \frac{8i^2}{n^3}=\frac{8}{n^3}\displaystyle\sum_{i=1}^{n} i^2[/latex].

Using the summation formula for [latex]\displaystyle\sum_{i=1}^{n} i^2[/latex], we have,

[latex]\begin{array}{ll} \displaystyle\sum_{i=1}^{n} f(x_i)\Delta x & =\frac{8}{n^3}\displaystyle\sum_{i=1}^{n} i^2 \\ & =\frac{8}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right] \\ & =\frac{8}{n^3}\left[\frac{2n^3+3n^2+n}{6}\right] \\ & =\frac{16n^3+24n^2+n}{6n^3} \\ & =\frac{8}{3}+\frac{4}{n}+\frac{1}{6n^2} \end{array}[/latex]

Now, to calculate the definite integral, we need to take the limit as [latex]n\to \infty[/latex]. We get,

[latex]\begin{array}{ll} \displaystyle\int_0^2 x^2 dx & =\underset{n\to \infty }{\lim} \displaystyle\sum_{i=1}^{n} f(x_i)\Delta x \\ & =\underset{n\to \infty }{\lim}\left(\frac{8}{3}+\frac{4}{n}+\frac{1}{6n^2}\right) \\ & =\underset{n\to \infty }{\lim}\left(\frac{8}{3}\right)+\underset{n\to \infty }{\lim}\left(\frac{4}{n}\right)+\underset{n\to \infty }{\lim}\left(\frac{1}{6n^2}\right) \\ & =\frac{8}{3}+0+0=\frac{8}{3} \end{array}[/latex]

Evaluating Definite Integrals

Evaluating definite integrals using Riemann sums can be quite tedious due to the complexity of the calculations. Later in this chapter, we will learn techniques for evaluating definite integrals without taking the limits of Riemann sums.

For now, we can rely on the fact that definite integrals represent the area under the curve. We can evaluate definite integrals by using geometric formulas to calculate that area. This helps us confirm that definite integrals do indeed represent areas, and we can then discuss how to handle cases where the curve of a function drops below the [latex]x[/latex]-axis.

Use the formula for the area of a circle to evaluate [latex]\displaystyle\int_3^6 \sqrt{9-(x-3)^2} dx[/latex].

Use the formula for the area of a trapezoid to evaluate [latex]\displaystyle\int_2^4 (2x+3) dx[/latex].