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Areas, Integrals, and Antiderivatives

Author: Sophia
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what's covered
In this lesson, you will connect the idea of area, definite integrals, and antiderivatives. Specifically, this lesson will cover:

Table of Contents

1. The Area Function, A (x) and Its Relationship to f  (x)

Consider the area of the region bounded by the t-axis (horizontal axis), the function f open parentheses t close parentheses equals 4 t comma and the vertical line t equals x. The graph is shown in the figure below.

A graph with an x-axis as t and a y-axis as f(x). A line slants upward from the point (0, 0) and passes through a marked point labeled ‘(x, 4x)’. A vertical dashed line starts downward from the marked point at (x, 4x) to meet the x-axis.

As the value of x changes, the area of the region changes, meaning that the area depends on x, meaning the area is a function of x.

Let A open parentheses x close parentheses equals the area of the region, which is a triangle.

Since A open parentheses x close parentheses defines the area of a region between f open parentheses t close parentheses and the t-axis, we can define A open parentheses x close parentheses as a definite integral:

A open parentheses x close parentheses equals integral subscript 0 superscript x 4 t d t

Assuming that x greater than 0 comma the area of the region is A open parentheses x close parentheses equals 1 half open parentheses x close parentheses open parentheses 4 x close parentheses equals 2 x squared.

try it
Consider the region bounded by f open parentheses t close parentheses equals 3 and the t-axis between t equals 0 and t equals x.
Write the area function A(x) as both a definite integral and as a function of x.
Here is a graph of f open parentheses t close parentheses equals 3 between t equals 0 and t equals x.



As you can see, the area of the region is 3 open parentheses x close parentheses equals 3 x.

If A open parentheses x close parentheses represents the area of the region, then A open parentheses x close parentheses equals 3 x.

Expressed as a definite integral, we write integral subscript 0 superscript x 3 d t.

Since these two quantities represent the same thing, we can say that integral subscript 0 superscript x 3 d t equals 3 x.

big idea
If x is one of the limits of integration, it is important that another variable be used inside the integral sign. This is because we can’t have variables serving two roles at once (upper limit of integration and the variable inside the integral).

EXAMPLE

Consider the region bounded by the t-axis and the line f open parentheses t close parentheses equals 2 plus 3 t between t equals 0 and t equals x.

A graph with an x-axis as t and a y-axis as f(x). A line slants upward from a marked point labeled ‘(0, 2)’ and crosses another marked point labeled ‘(x, 2 + 3x)’. A vertical dashed line starts downward from the point (x, 2 + 3x) to meet the x-axis. Another dashed line overlaps the y-axis between (0, 0) and (0, 2).

As x increases, the shape remains a trapezoid.

As a definite integral, the area is:
A open parentheses x close parentheses equals integral subscript 0 superscript x open parentheses 2 plus 3 t close parentheses d t

Using the trapezoid area formula, we have:

1 half open parentheses x close parentheses open parentheses 2 plus open parentheses 2 plus 3 x close parentheses close parentheses Use the trapezoid area formula.
1 half x open parentheses 4 plus 3 x close parentheses Simplify parentheses.
2 x plus 3 over 2 x squared Distribute.

Thus, A open parentheses x close parentheses equals 2 x plus 3 over 2 x squared.

You might notice that there is a relationship between the area function A open parentheses x close parentheses and the associated curve y equals f open parentheses x close parentheses. We’re going to explore this in this next segment.

Consider the last three examples. Here is a summary of the area functions with their associated curves, as well as the derivatives of each area function.

Regions Area Function, bold italic A open parentheses bold x close parentheses “Height” Function, bold italic f open parentheses bold x close parentheses bold italic A bold apostrophe open parentheses bold x close parentheses
Region bounded by the t-axis (horizontal axis), the function f open parentheses t close parentheses equals 4 t comma and the vertical line t equals x A open parentheses x close parentheses equals 2 x squared f open parentheses x close parentheses equals 4 x A apostrophe open parentheses x close parentheses equals 4 x
Region bounded by f open parentheses t close parentheses equals 3 and the t-axis between t equals 0 and t equals x A open parentheses x close parentheses equals 3 x f open parentheses x close parentheses equals 3 A apostrophe open parentheses x close parentheses equals 3
Region bounded by the t-axis and the line f open parentheses t close parentheses equals 2 plus 3 t between t equals 0 and t equals x A open parentheses x close parentheses equals 2 x plus 3 over 2 x squared f open parentheses x close parentheses equals 2 plus 3 x A apostrophe open parentheses x close parentheses equals 2 plus 3 x

Note that in each situation, A apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses. It turns out that this is always the case, which is a very useful idea in finding areas of regions that use any choice of f open parentheses x close parentheses. First, we need to learn a bit about antiderivatives.

2. Finding Basic Antiderivatives

We call F open parentheses x close parentheses an antiderivative of f open parentheses x close parentheses if F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses. That is, F open parentheses x close parentheses is the function whose derivative is f open parentheses x close parentheses. For instance, an antiderivative of f open parentheses x close parentheses equals 3 x squared is F open parentheses x close parentheses equals x cubed since D open square brackets x cubed close square brackets equals 3 x squared. In fact, we could also say that F open parentheses x close parentheses equals x cubed plus 4 is an antiderivative of f open parentheses x close parentheses equals 3 x squared since D open square brackets x cubed plus 4 close square brackets equals 3 x squared.

As it turns out, any function of the form F open parentheses x close parentheses equals x cubed plus C (where C is constant) is an antiderivative of f open parentheses x close parentheses equals 3 x squared since D open square brackets x cubed plus C close square brackets equals 3 x squared.

EXAMPLE

Find three antiderivatives of f open parentheses x close parentheses equals cos x.

Since D open square brackets sin x close square brackets equals cos x comma it follows that F open parentheses x close parentheses equals sin x is an antiderivative of f open parentheses x close parentheses equals cos x. To find others, all you would need to do is add a constant.

Two more antiderivatives are F open parentheses x close parentheses equals sin x plus 2 and F open parentheses x close parentheses equals sin x minus 3.

In summary, any function of the form F open parentheses x close parentheses equals sin x plus C is an antiderivative of f open parentheses x close parentheses equals cos x.

try it
Consider the function f open parentheses x close parentheses equals 6 x squared.
Write three antiderivatives of this function.
Any function of the form F open parentheses x close parentheses equals 2 x cubed plus C is an antiderivative of f open parentheses x close parentheses equals 6 x squared. Here are three examples:
  • F open parentheses x close parentheses equals 2 x cubed
  • F open parentheses x close parentheses equals 2 x cubed plus 2
  • F open parentheses x close parentheses equals 2 x cubed minus 4

term to know
Antiderivative
F open parentheses x close parentheses is an antiderivative of f open parentheses x close parentheses if F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses.

3. The First and Second Fundamental Theorems of Calculus

3a. The First Fundamental Theorem of Calculus

Consider the area function A open parentheses x close parentheses equals integral subscript 0 superscript x f open parentheses t close parentheses d t.

By substituting x equals 1 and x equals 2 comma we have A open parentheses 1 close parentheses equals integral subscript 0 superscript 1 f open parentheses t close parentheses d t and A open parentheses 2 close parentheses equals integral subscript 0 superscript 2 f open parentheses t close parentheses d t.

By properties of integrals, integral subscript 0 superscript 2 f open parentheses t close parentheses d t equals integral subscript 0 superscript 1 f open parentheses t close parentheses d t plus integral subscript 1 superscript 2 f open parentheses t close parentheses d t.

Replacing the first two integrals by their values, we have A open parentheses 2 close parentheses equals A open parentheses 1 close parentheses plus integral subscript 1 superscript 2 f open parentheses t close parentheses d t.

Finally, let's write the definite integral to one side: A open parentheses 2 close parentheses minus A open parentheses 1 close parentheses equals integral subscript 1 superscript 2 f open parentheses t close parentheses d t

Remember that A apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses comma meaning that A open parentheses x close parentheses is an antiderivative of f open parentheses x close parentheses. Therefore, we could write A open parentheses x close parentheses equals F open parentheses x close parentheses.

Thus, we can rewrite as integral subscript 1 superscript 2 f open parentheses t close parentheses d t equals F open parentheses 2 close parentheses minus F open parentheses 1 close parentheses. This is generalized in the first fundamental theorem of calculus, as shown below:

Let F open parentheses x close parentheses be an antiderivative of f open parentheses x close parentheses comma meaning that F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses. Then, integral subscript a superscript b f open parentheses x close parentheses d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses comma which means we evaluate the antiderivative at the endpoints, then subtract.

To show that we are substituting a and b into F open parentheses x close parentheses comma we use the following notation: F open parentheses x close parentheses open vertical bar blank subscript a superscript b close

Then, it follows that F open parentheses x close parentheses open vertical bar equals presubscript a presuperscript b F open parentheses b close parentheses close minus F open parentheses a close parentheses.

EXAMPLE

Evaluate integral subscript 0 superscript 2 3 x squared d x.

Since any function of the form F open parentheses x close parentheses equals x cubed plus C is an antiderivative of f open parentheses x close parentheses equals 3 x squared comma we have the following:

integral subscript 0 superscript 2 3 x squared d x Start with the original expression.
equals open parentheses x cubed plus C close parentheses open vertical bar blank subscript 0 superscript 2 close Apply the first fundamental theorem of calculus with F open parentheses x close parentheses equals x cubed plus C.
equals open parentheses 2 cubed plus C close parentheses minus open parentheses 0 cubed plus C close parentheses Substitute x equals 2 and x equals 0 into F open parentheses x close parentheses comma then subtract.
equals 8 plus C minus C Evaluate operations in the parentheses.
equals 8 Simplify.

Thus, integral subscript 0 superscript 2 3 x squared d x equals 8.

big idea
Notice that the “+C” dropped out when evaluating the definite integral. Intuitively, this will always happen. Thus, when evaluating a definite integral, select C equals 0. Some students even say “You don’t need the C.”

term to know
The First Fundamental Theorem of Calculus
Let F open parentheses x close parentheses be an antiderivative of f open parentheses x close parentheses comma meaning that F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses.
Then, integral subscript a superscript b f open parentheses x close parentheses d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses comma which means we evaluate the antiderivative at the endpoints, then subtract.

3b. The Second Fundamental Theorem of Calculus

Recall that A apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses. If we replace A open parentheses x close parentheses with F open parentheses x close parentheses to correspond with f open parentheses x close parentheses comma we have another important theorem in calculus, the second fundamental theorem of calculus:

Let f open parentheses x close parentheses be a continuous function on the closed interval open square brackets a comma space b close square brackets with a less or equal than x less or equal than b.

Let F open parentheses x close parentheses equals integral subscript a superscript x f open parentheses t close parentheses d t. Then, F apostrophe open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction integral subscript a superscript x f open parentheses t close parentheses d t equals f open parentheses x close parentheses.

EXAMPLE

Let F open parentheses x close parentheses equals integral subscript 2 superscript x square root of t plus 1 end root d t.

Since f open parentheses t close parentheses equals square root of t plus 1 end root is continuous on open square brackets short dash 1 comma space infinity close parentheses comma which includes 2, then F apostrophe open parentheses x close parentheses equals square root of x plus 1 end root.

try it
Let F open parentheses x close parentheses equals integral subscript square root of straight pi end subscript superscript x sin open parentheses t squared close parentheses d t.
Find F '(x).
Since square root of straight pi is a constant, the 2nd fundamental theorem of calculus can be used directly, which means the derivative is simply the integrand with t replaced with x.

That is, F apostrophe open parentheses x close parentheses equals sin open parentheses x squared close parentheses.

Suppose x is replaced by u, where u is a function of x.

That is, F open parentheses x close parentheses equals integral subscript a superscript u f open parentheses t close parentheses d t. Then, by the chain rule, F apostrophe open parentheses x close parentheses equals f open parentheses u close parentheses times fraction numerator d u over denominator d x end fraction.

EXAMPLE

Let F open parentheses x close parentheses equals integral subscript 1 superscript x cubed end superscript e to the power of t d t.

Then, F apostrophe open parentheses x close parentheses equals e to the power of x cubed end exponent times 3 x squared equals 3 x squared e to the power of x cubed end exponent.

try it
Let F open parentheses x close parentheses equals integral subscript 0 superscript sin x end superscript ln t d t.
Find F '(x).
From the second fundamental theorem of calculus:

F apostrophe open parentheses x close parentheses equals ln open parentheses sin x close parentheses times D open square brackets sin x close square brackets</dd></dl> space space space space space space space space space equals ln open parentheses sin x close parentheses times cos x space space space space space space space space space equals cos x times ln open parentheses sin x close parentheses

term to know
The Second Fundamental Theorem of Calculus
Let f open parentheses x close parentheses be a continuous function on the closed interval open square brackets a comma space b close square brackets with a less or equal than x less or equal than b.
Let F open parentheses x close parentheses equals integral subscript a superscript x f open parentheses t close parentheses d t. Then, F apostrophe open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction integral subscript a superscript x f open parentheses t close parentheses d t equals f open parentheses x close parentheses.

4. Using Antiderivatives to Calculate Area

As a result of the fundamental theorem of calculus, we have a new way to compute areas with definite integrals. Instead of relying on a sketch of the region, we can use antiderivatives to compute areas.

EXAMPLE

Find the area between the graph of f open parentheses x close parentheses equals cos x and the x-axis between x equals 0 and x equals straight pi over 2. The region is shown in the figure:

A graph with an x-axis ranging from 0 to 1.5 at intervals of 0.5 and a y-axis ranging from 0 to 1 at intervals of 0.2. A curve starts from the upper left part of the y-axis, reaches a peak at (0, 1), and then dips downward by crossing the y-axis at (0, 1). From here, the curve extends down by crossing the x-axis at (1.51, 0). The area under the curve from the origin (0, 0) to x equals 1.51 is shaded.

We know the area is given by the definite integral integral subscript 0 superscript straight pi divided by 2 end superscript cos x d x.

Earlier, we saw that sin x is an antiderivative of cos x. Therefore, the area is as follows:

integral subscript 0 superscript straight pi divided by 2 end superscript cos x d x Start with the original expression.
equals sin x open vertical bar blank subscript 0 superscript straight pi divided by 2 end superscript close Use the fundamental theorem of calculus with F open parentheses x close parentheses equals sin x. Remember, we do not need to write “+C,” meaning we are choosing C equals 0.
equals sin straight pi over 2 minus sin 0 Substitute x equals straight pi over 2 and x equals 0 into F open parentheses x close parentheses comma then subtract.
equals 1 Simplify. Recall sin straight pi over 2 equals 1 and sin 0 equals 0.

Thus, the area of the region is 1 square unit.

try it
Consider the graphs of f open parentheses x close parentheses equals e to the power of x and the x-axis between x equals 0 and x equals 2.
Calculate the area between these graphs.
The graph of the region is shown below.



The definite integral used to find the area is integral subscript 0 superscript 2 e to the power of x d x.

Since the antiderivative of e to the power of x is e to the power of x comma we have:

integral subscript 0 superscript 2 e to the power of x d x equals e to the power of x open vertical bar equals presubscript 0 presuperscript 2 e squared close minus e to the power of 0 equals e squared minus 1

summary
In this lesson, you learned that with a link between the area function bold italic A bold left parenthesis bold italic x bold right parenthesis and a region with bold italic f bold left parenthesis bold italic x bold right parenthesis as the upper boundary, basic antiderivatives can be used to calculate areas and compute definite integrals by using the first and second fundamental theorems of calculus.

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
Antiderivative

F open parentheses x close parentheses is an antiderivative of f open parentheses x close parentheses if F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses.

The First Fundamental Theorem of Calculus

Let F open parentheses x close parentheses be an antiderivative of f open parentheses x close parentheses comma meaning that F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses.

Then, integral subscript a superscript b f open parentheses x close parentheses d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses comma which means we evaluate the antiderivative at the endpoints, then subtract.

The Second Fundamental Theorem of Calculus

Let f open parentheses x close parentheses be a continuous function on the closed interval open square brackets a comma space b close square brackets with a less or equal than x less or equal than b.

Let F open parentheses x close parentheses equals integral subscript a superscript x f open parentheses t close parentheses d t. Then, F apostrophe open parentheses x close parentheses equals fraction numerator d over denominator d x end fraction integral subscript a superscript x f open parentheses t close parentheses d t equals f open parentheses x close parentheses.

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