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Definition of the Definite Integral

Author: Sophia
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what's covered
In this lesson, you will connect Riemann sums and the definition of the definite integral. This is the key step in moving into integral calculus. Specifically, this lesson will cover:

Table of Contents

1. The Definition of the Definite Integral

Consider the area of the region bounded by y equals x squared plus 2 between the x-axis, x equals 1 comma and x equals 3.

Shown below is the actual region as well as the region approximated by 4, 8, and 16 rectangles; all have equal width.

Actual Region 4 Rectangles
A graph with an x-axis ranging from 1 to 3. A curve opens upward, starts from the marked point (1, 3), and rises to the marked point (3, 11). The area under the curve and above the horizontal axis between the marked points (1, 3) and (3, 11) is shaded. A graph with an x-axis ranging from 1 to 3 at intervals of 0.5. Four rectangular bars correspond to subintervals 1–1.5, 1.5–2, 2–2.5, and 2.5–3, respectively, increasing in height as x increases. A curve rises upward from the left corresponding to x equals 1, touches the top left corner of each bar, and rises to a point corresponding to x equals 3, forming an ascending stair-like pattern. The total area of the rectangular bars is shaded.
8 Rectangles 16 Rectangles
A graph with an x-axis ranging from 1 to 3. Eight rectangular bars correspond to subintervals 1.0−1.25, 1.25−1.5, 1.5−1.75, 1.75−2.0, 2.0−2.25, 2.5−2.75, and 2.75−3.0, respectively, on the x-axis, increasing in height as x increases. A curve rises upward from the first bar corresponding to x equals 1, touches the top left corner of each bar, and rises to a point corresponding to x equals 3. The total area of the rectangular bars is shaded. A graph with an x-axis ranging from 1 to 3. 16 rectangular bars correspond to subintervals 1.0−1.125, 1.125−1.25, 1.25−1.375, 1.375−1.5, 1.5−1.625, 1.625−1.75, 1.75−1.875, 1.875−2.0, 2.0−2.125, 2.125−2.25, 2.25−2.375, 2.375−2.5, 2.5−2.625, 2.625−2.75, 2.75−2.875, and 2.875−3.0, respectively, on the x-axis, increasing in height as x increases. A curve rises upward from the first bar corresponding to x equals 1, touches the top left corner of each bar, and rises to a point corresponding to x equals 3. The total area of the rectangular bars is shaded.

When the subintervals have equal width, we notice the following as the number of rectangles (and subintervals) increases:

  1. The width of each rectangle (and subinterval) decreases.
  2. The sum of the areas of the rectangles gets closer to the actual area under the curve.
When calculating a Riemann sum for a function f open parentheses x close parentheses on open square brackets a comma space b close square brackets comma we will only use rectangles that have equal width. That is, when n subintervals are used, the width of each subinterval is increment x equals fraction numerator b minus a over denominator n end fraction. This also means that Riemann sums from this point forward will be written as:

sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x

Assume (for now) that f open parentheses x close parentheses is nonnegative. As the number of rectangles gets larger, which means that n rightwards arrow infinity comma this quantity will get closer to the actual area as long as increment x rightwards arrow 0 for all k.

When limit as n rightwards arrow infinity of sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x exists and has the same value regardless of the values of c subscript k used in each subinterval, then f open parentheses x close parentheses is integrable on open square brackets a comma space b close square brackets.

If we call A the definite integral of f open parentheses x close parentheses on open square brackets a comma space b close square brackets comma then A equals integral subscript a superscript b f open parentheses x close parentheses d x.

In this notation,

  • The numbers a and b represent the lower and upper limits of integration.
  • The function f open parentheses x close parentheses is called the integrand.
  • x is called the variable of integration.
  • The differential dx tells us that the definite integral is computed by letting values of x increase from a to b.
big idea
For a non-negative function f open parentheses x close parentheses comma the value of the Riemann sum approaches the definite integral as n rightwards arrow infinity. Then, the quantity integral subscript a superscript b f open parentheses x close parentheses d x is the area between the graph of a nonnegative function f open parentheses x close parentheses and the x-axis, between x equals a and x equals b.

term to know
Integrable
If the value of limit as n rightwards arrow infinity of sum from k equals 1 to n of f open parentheses c subscript k close parentheses times increment x exists and is equal to A regardless of the values of c subscript k used in each subinterval, then we say that f open parentheses x close parentheses is integrable on the interval open square brackets a comma space b close square brackets.

2. Definite Integrals and Riemann Sums

Using the definition of a definite integral, we can write Riemann sums as definite integrals and vice versa.

EXAMPLE

Write integral subscript 0 superscript 4 2 x d x as a Riemann sum.

Recall the Riemann sum for a function f open parentheses x close parentheses is sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x.

Since f open parentheses x close parentheses equals 2 x comma the definite integral is the value of the Riemann sum as n rightwards arrow infinity colon space limit as n rightwards arrow infinity of sum from k equals 1 to n of 2 c subscript k increment x

Now, let’s take a Riemann sum and write it as a definite integral.

EXAMPLE

A function f open parentheses x close parentheses on the interval open square brackets short dash 4 comma space 4 close square brackets has the Riemann sum sum from k equals 1 to n of fraction numerator 1 over denominator 1 plus c subscript k superscript 2 end fraction increment x.

The definite integral is the value of the Riemann sum as n rightwards arrow infinity comma and is written integral subscript short dash 4 end subscript superscript 4 fraction numerator 1 over denominator 1 plus x squared end fraction d x.

try it
Given below are (a) the limit of a Riemann sum and (b) a definite integral.
(a) limit as n rightwards arrow infinity of sum from k equals 1 to n of sin open parentheses c subscript k close parentheses increment x comma space 0 less or equal than x less or equal than straight pi
(b) integral subscript 0 superscript 9 square root of x plus 16 end root d x
Write the definite integral that corresponds to the Riemann sum in (a).
integral subscript 0 superscript straight pi sin x d x
Write the limit of the Riemann sum that corresponds to the definite integral in (b).
limit as n rightwards arrow infinity of sum from k equals 1 to n of square root of c subscript k plus 16 end root increment x comma space 0 less or equal than x less or equal than 9

3. Using Riemann Sums to Evaluate Definite Integrals

We learned that f open parentheses x close parentheses is integrable on open square brackets a comma space b close square brackets if limit as n rightwards arrow infinity of sum from k equals 1 to n of f open parentheses c subscript k close parentheses times increment x exists and is equal to the same value for any choice of c subscript 1 comma space c subscript 2 comma space horizontal ellipsis comma space c subscript n comma each of which is in their respective subintervals.

By using the formulas for sigma notation combined with the limit definition, we can evaluate some definite integrals of functions for which we don’t know the area of the corresponding region.

Here is how:

step by step
  1. Since each subinterval has equal width, we know increment x equals fraction numerator b minus a over denominator n end fraction.
  2. Select c subscript k to be the right-hand endpoint of the interval. Then, c subscript 1 equals a plus increment x comma c subscript 2 equals a plus 2 increment x comma c subscript 3 equals a plus 3 increment x comma ... which means c subscript k equals a plus k increment x.
  3. Substitute c subscript k into the function.
  4. Evaluate the sum (using formulas), then compute the limit.

EXAMPLE

Use a Riemann sum to evaluate integral subscript 0 superscript 4 2 x d x.
  1. Find the width of each subinterval: increment x equals fraction numerator 4 minus 0 over denominator n end fraction equals 4 over n
  2. Find the right-hand endpoints: c subscript k equals a plus k increment x equals 0 plus k open parentheses 4 over n close parentheses equals fraction numerator 4 k over denominator n end fraction
  3. Evaluate the function at each c subscript k colon f open parentheses c subscript k close parentheses equals 2 open parentheses fraction numerator 4 k over denominator n end fraction close parentheses equals fraction numerator 8 k over denominator n end fraction
  4. Then, integral subscript 0 superscript 4 2 x d x equals limit as n rightwards arrow infinity of sum from k equals 1 to n of open parentheses fraction numerator 8 k over denominator n end fraction close parentheses open parentheses 4 over n close parentheses.

Next, simplify the sum and calculate the limit.

limit as n rightwards arrow infinity of sum from k equals 1 to n of open parentheses fraction numerator 8 k over denominator n end fraction close parentheses open parentheses 4 over n close parentheses Calculate the limit.
equals limit as n rightwards arrow infinity of sum from k equals 1 to n of open parentheses fraction numerator 32 k over denominator n squared end fraction close parentheses Simplify.
equals limit as n rightwards arrow infinity of 32 over n squared sum from k equals 1 to n of k 32 over n squared is a constant factor since k is the index of summation. Therefore, it can be factored out.
equals limit as n rightwards arrow infinity of open parentheses 32 over n squared times fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close parentheses Apply the summation formula: sum from k equals 1 to n of k equals fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction
equals limit as n rightwards arrow infinity of fraction numerator 16 n squared plus 16 n over denominator n squared end fraction Simplify.
equals limit as n rightwards arrow infinity of open parentheses 16 plus 16 over n close parentheses Divide each term by n squared.
equals 16 Evaluate the limit.

Regardless of the values of c subscript k used (left endpoints, etc.), this holds true.

Thus, integral subscript 0 superscript 4 2 x d x equals 16.

watch
In this video, we’ll use the Riemann sum to evaluate integral subscript 1 superscript 3 open parentheses x squared plus 2 close parentheses d x.

try it
Consider the integral integral subscript 2 superscript 6 open parentheses 10 minus x close parentheses d x.
Use the Riemann sum to evaluate this integral.
Since the integral is taken from x equals 2 to x equals 6 comma we need to set up the values of c subscript k.

When using n subintervals, the width of each rectangle is increment x equals fraction numerator 6 minus 2 over denominator n end fraction equals 4 over n.

Then, c subscript k equals a plus k increment x equals 2 plus k open parentheses 4 over n close parentheses equals 2 plus fraction numerator 4 k over denominator n end fraction

Then, the Riemann Sum for n rectangles is:

sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x equals sum from k equals 1 to n of open square brackets 10 minus open parentheses 2 plus fraction numerator 4 k over denominator n end fraction close parentheses close square brackets open parentheses 4 over n close parentheses equals sum from k equals 1 to n of open square brackets 8 minus fraction numerator 4 k over denominator n end fraction close square brackets open parentheses 4 over n close parentheses

Multiplying and simplifying, we have the following (note that n is a constant since the summation depends on k).

equals sum from k equals 1 to n of open parentheses 32 over n minus fraction numerator 16 k over denominator n squared end fraction close parentheses equals sum from k equals 1 to n of 32 over n minus 16 over n squared sum from k equals 1 to n of k

Evaluating each sum, we have:

equals n open parentheses 32 over n close parentheses minus 16 over n squared open parentheses fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close parentheses equals 32 minus fraction numerator 8 open parentheses n plus 1 close parentheses over denominator n end fraction

Lastly, the value of the definite integral is the limit of this expression as n rightwards arrow infinity comma we have:

limit as n rightwards arrow infinity of open square brackets 32 minus fraction numerator 8 open parentheses n plus 1 close parentheses over denominator n end fraction close square brackets equals limit as n rightwards arrow infinity of 32 minus limit as n rightwards arrow infinity of fraction numerator 8 open parentheses n plus 1 close parentheses over denominator n end fraction equals 32 minus 8 equals 24

Thus, the value of the definite integral is 24.

think about it
Fortunately, this is not the only method to evaluate definite integrals. These samples were chosen based on known summation formulas.

For example, consider the definite integral integral subscript 0 superscript straight pi sin x d x. The corresponding limit of a Riemann sum is limit as n rightwards arrow infinity of sum from k equals 1 to n of open square brackets sin open parentheses fraction numerator k straight pi over denominator n end fraction close parentheses times straight pi over n close square brackets comma which has no known summation formula. We will learn how to find the value of the definite integral without summations in a future challenge.

4. Using Area to Evaluate Riemann Sums and Definite Integrals

EXAMPLE

Evaluate the definite integral: integral subscript 0 superscript 4 2 x d x

The figure shows the region bounded by the graph of f open parentheses x close parentheses equals 2 x and the x-axis on the interval open square brackets 0 comma space 4 close square brackets.

A graph with an x-axis ranging from 0 to 4 and a y-axis ranging from 0 to 8 at intervals of 2. A line slants upward from a marked point at (0, 0) to a marked point at (4, 8). The area below the line and above the horizontal axis is shaded.

The region is the shape of a triangle with base 4 and height 8, which has area A equals 1 half open parentheses 4 close parentheses open parentheses 8 close parentheses equals 16 square units.

Then, integral subscript 0 superscript 4 2 x d x equals 16.

Let’s look at an example of a continuous piecewise function.

EXAMPLE

Consider the graph of f open parentheses x close parentheses shown in the figure. Use it to evaluate integral subscript 0 superscript 6 f open parentheses x close parentheses d x.

A graph with an x-axis ranging from 0 to 6 at intervals of 0.5 and a y-axis ranging from 0 to 5. A line slants downward from a marked point at (0, 4) to the marked point at (4, 0) and then slants upward to the marked point at (6, 2), where it stops.

The region is shown in the figure below and is composed of two triangles.

A graph with an x-axis ranging from 0 to 6 at intervals of 0.5 and a y-axis ranging from 0 to 5. A line slants downward from a marked point at (0, 4) to the marked point at (4, 0) and then slants upward to the marked point at (6, 2), where it stops. The area below the lines and above the axis is shaded.

The triangle on open square brackets 0 comma space 4 close square brackets has area 1 half open parentheses 4 close parentheses open parentheses 4 close parentheses equals 8 space units squared and the triangle on open square brackets 4 comma space 6 close square brackets has area 1 half open parentheses 2 close parentheses open parentheses 2 close parentheses equals 2 space units squared.

The total area is 10 space units squared comma which means that integral subscript 0 superscript 6 f open parentheses x close parentheses d x equals 10.

try it
Consider the graph of f open parentheses x close parentheses shown in the figure that can be used to evaluate integral subscript short dash 2 end subscript superscript 2 f open parentheses x close parentheses d x.

A graph with an x-axis ranging from –2 to 2 and a y-axis ranging from 0 to 8. A line slants downward from a marked point (–2, 8) in the second quadrant to reach a marked point at (0, 6) and then extends downward to the marked point at (2, 0) on the x-axis. 

Evaluate this integral.
The value of the definite integral is the total area between the graph and the x-axis.

A graph with an x-axis ranging from −2 to 2 and a y-axis ranging from 0 to 8. A line starts from the marked point at (−2, 8) in the second quadrant and extends to the marked point at (0, 6). From here, it continues downward to the marked point at (2, 0). The area below the line between x equals −2 and x equals 0 is shaded, representing a trapezoid, and the area below the line between x equals 0 and x equals 2 is also shaded, representing a triangle.

From x equals short dash 2 to x equals 0 comma the region is a trapezoid with bases 8 and 6, and height 2. The area of this region is A subscript 1 equals 2 over 2 open parentheses 8 plus 6 close parentheses equals 14 space units squared.

From x equals 0 to x equals 2 comma the region is a triangle with base 2 and height 6. The area of this region is A subscript 2 equals 1 half open parentheses 2 close parentheses open parentheses 6 close parentheses equals 6 space units squared.

Then, the value of integral subscript short dash 2 end subscript superscript 2 f open parentheses x close parentheses d x equals A subscript 1 plus A subscript 2 equals 14 plus 6 equals 20 space units squared.

As it turns out, f open parentheses x close parentheses doesn't have to be continuous in order to be integrable. Here is an example that illustrates this.

watch
In this video, we’ll use a graph of some function y equals f open parentheses x close parentheses shown to evaluate integral subscript 0 superscript 10 f open parentheses x close parentheses d x.

summary
In this lesson, you learned the definition of the definite integral, understanding that for a non-negative function f open parentheses x close parentheses comma the value of the Riemann sum approaches the definite integral as n rightwards arrow infinity. You also learned that by using the formulas for sigma notation combined with the limit definition, you can evaluate definite integrals of functions for which you don’t know the area of the corresponding region, by using Riemann sums to visualize how the area between f open parentheses x close parentheses and the x-axis on open square brackets a comma space b close square brackets is obtained. Finally, through a series of examples using area to evaluate Riemann sums and definite integrals, you have seen that f open parentheses x close parentheses does not have to be continuous to be integrable.

Source: THIS TUTORIAL HAS BEEN ADAPTED FROM CHAPTER 4 OF "CONTEMPORARY CALCULUS" BY DALE HOFFMAN. ACCESS FOR FREE AT WWW.CONTEMPORARYCALCULUS.COM. LICENSE: CREATIVE COMMONS ATTRIBUTION 3.0 UNITED STATES.

Terms to Know
Integrable

If the value of limit as n rightwards arrow infinity of sum from k equals 1 to n of f open parentheses c subscript k close parentheses times increment x exists and is equal to A regardless of the values of c subscript k used in each subinterval, then we say that f open parentheses x close parentheses is integrable on the interval open square brackets a comma space b close square brackets.

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