Fundamental Theorem Of Calculus Essay

Introduction

The Fundamental Theorem of Calculus brings together two essential concepts in calculus: differentiation and integration. There are two parts to the Fundamental Theorem: the first justifies the procedure for evaluating definite integrals, and the second establishes the relationship between differentiation and integration. Here, we will focus on the first statement, which is referred to as the First Fundamental Theorem of Calculus.

This theorem establishes the procedure for computing a definite integral. Recall that a definite integral is an integral where you are given the limits of integration. That is, you are integrating over an interval whose endpoints you use to evaluate the integral.

Indefinite vs. Definite Integrals

An indefinite integral is an integral without limits of integration; for example,

\int { { x }^{ 2 } } dx

When integrating this function, we are looking for a curve whose derivative is {x}^{2}. There are infinitely many curves with that derivative, including

\frac { 1 }{ 3 } { x }^{ 3 }

\frac { 1 }{ 3 } { x }^{ 3 }+1

\frac { 1 }{ 3 } { x }^{ 3 }+8

\frac { 1 }{ 3 } { x }^{ 3 }-3

\frac { 1 }{ 3 } { x }^{ 3 }-\sqrt { 5 }

While there are infinitely many options, we note that they differ only by a constant. As a result, when we integrate a function of this type, we add an arbitrary constant, C, to our solution, as follows:

\int { { x }^{ 2 } } =\frac { 1 }{ 3 } { x }^{ 3 }+C

In contrast, a definite integral is one in which there are limits of integration. As an example, consider \int _{ 1 }^{ 4 }{ { x }^{ 2 } dx } . Here, we are being asked to integrate the same function as before, but we are doing so over the interval 1\le x\le 4. To compute this integral, we rely on the First Fundamental Theorem of Calculus.

The First Fundamental Theorem of Calculus

The theorem states that if a function f(x) has an indefinite integral of F(x)+C and is continuous over an interval [a,b], then \int _{ a }^{ b }{ f(x)dx=F(b)-F(a) }.

Let’s go back to our prior example:

\int _{ 1 }^{ 4 }{ { x }^{ 2 } dx }

First, we integrate { x }^{ 2 } to obtain \frac { 1 }{ 3 } { x }^{ 3 } . When evaluating definite integrals, we leave out the constant (so there is no +C here).

Next, we plug the limits of integration into the integral we just found. We first evaluate the upper limit of integration (F(b)), then subtract the value of the function at the lower limit of integration (F(a)). Putting this all together, we obtain the following expression:

\int _{ 1 }^{ 4 }{ { x }^{ 2 } dx } ={ \left[ \frac { 1 }{ 3 } { x }^{ 3 } \right] }_{ 1 }^{ 4 }=\frac { 1 }{ 3 } { (4) }^{ 3 }-\frac { 1 }{ 3 } { (1) }^{ 3 }=\frac { 1 }{ 3 } (64-1)=\frac { 1 }{ 3 } (63)=21

Why does it Work?

Let G(x)=\int _{ a }^{ x }{ f(t)dt } .

We can show that F(x) is an antiderivative for f on some interval [a,b]. This result is actually because of the Second Fundamental Theorem of Calculus. We can show that the second fundamental theorem works without using the first, but interestingly, proving the first part is easiest if we start with the second part.

The two anti-derivatives are continuous over the given interval. Also, { G }^{ ' }(x)={ F }^{ ' }(x)=f(x) for all x in the interval [a,b] . In other words, since G(x) and F(x) are anti-derivatives of f(x), their derivative is f(x).

You might remember that a function has multiple anti-derivatives and that these differ by a constant. This is why you add +C at the end of your answers when you evaluate an indefinite integral (one without limits of integration). As a result, we know there exists another anti-derivative of f on some interval [a,b]. Let us call this anti-derivative F(x), where G(x)=F(x)+C.

Recall that G(a)=\int _{ a }^{ a }{ f(t)dt=0 }. That is, the integral at a particular point within the interval is equal to zero. From this statement, we arrive at the following:

0=G(a)=F(a)+C

C=-F(a)

We noted before that G(x)=F(x)+C. Since we just found a value for C, we have

G(x)=F(b)-F(a).

If we let x=b, we arrive at G(b)=F(b)-F(a). Using the relationship between anti-derivatives and derivatives, we conclude that

\int _{ a }^{ b }{ f(t)dt=F(b)-F(a) }.

Example 1:

Find \int _{ 0 }^{ \pi /2 }{ cos(x)dx }.

In this straightforward example, we are asked to evaluate a definite integral. Recall that the derivative of \sin { x } is \cos { x }, so the anti-derivative of \cos { x } is \sin { x } + C. However, as this is a definite integral (i.e. we are given the limits of integration), we need not add the constant (C). We evaluate the integral by using the first fundamental theorem as follows:

\int _{ 0 }^{ \pi /2 }{ \cos { x }dx } ={ \left[ \sin { x } \right] }_{ 0 }^{ \pi /2 }=\ sin { (\pi /2) } - \sin { (0) } =1-0=1

Example 2: (1988 AP Calculus Exam Multiple Choice Problem 10)

If \int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx=18 } what is the value of k?

Here, we have removed the multiple choice answers, but the question is identical to one that appeared on the AP exam in all other respects.

We are asked to find k, which involves evaluating the given definite integral. Invoking the first fundamental theorem of calculus, we know that we must find the difference between the values of the anti-derivative at k and at 0.

Keep in mind that k is a constant (a number), as this will help us with the anti-derivative. Specifically, when finding the anti-derivative of 2kx, we treat k as we might the number 2. That is, we leave it alone! It is not a variable and so there should be no \frac { { k }^{ 2 } }{ 2 } anywhere in your work!

We find the anti-derivative as follows:

\int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx } ={ \left[ \frac { 2k{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ k }={ \left[ k{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ k }={ \left[ \frac { 3k{ x }^{ 2 } }{ 3 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ k }={ \left[ \frac { 3k{ x }^{ 2 }-{ x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ k }

Now, we apply the Fundamental Theorem of Calculus. This means that we evaluate the definite integral by using the limits of integration (first substituting k for x, then 0 for x) as follows.

{ \left[ \frac { 3k{ x }^{ 2 }-{ x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ k }=\frac { 3k\cdot { k }^{ 2 }-{ k }^{ 3 } }{ 3 } -\frac { 3k{ (0) }^{ 2 }-{ (0) }^{ 3 } }{ 3 } =\frac { 3{ k }^{ 3 }-{ k }^{ 3 } }{ 3 } -0=\frac { 2{ k }^{ 3 } }{ 3 }

Recall that

\int _{ 0 }^{ k }{ (2kx-{ x }^{ 2 })dx=18 }. Thus, \frac { 2{ k }^{ 3 } }{ 3 } =18 and { k }^{ 3 }=\left( 18 \right) \frac { 3 }{ 2 } =27

k=3

Example 3: (1988 AP Calculus Exam Multiple Choice Problem 13)

If the function f has a continuous derivative on [0,C] then \int _{ 0 }^{ c }{ { f }^{ ' }(x)dx=? }

  1. f(c)-f(0)
  2. \left| f(c)-f(0) \right|
  3. f(c)
  4. f(x)+C
  5. { f }^{ " }(c)-{ f }^{ " }(0)

This problem is testing us on our understanding of the Fundamental Theorem of Calculus. The anti-derivative of the derivative of f is f itself. Next, we evaluate the definite integral by using the limits of integration. The answer is (A).

What does \int _{ a }^{ b }{ f(t)dt=F(b)-F(a) } actually mean?

We can understand the integral \int _{ a }^{ b }{ f(t) } dt in a number of ways. One of these is to think of the integral as the total change in the curve over the interval from a to b. The total change is the definite integral of the rate of change of a function. Below are some questions from actual AP Calculus exams that require the use of the First Fundamental Theorem of Calculus and the fact that the definite integral gives the total change of the function over the given integral.

Example: (2007 AP Calculus AB Exam Question 2 part b)

Image 1: The CollegeBoard Advanced Placement Program

We will focus on part b and note that we are asked for the total distance that the particle traveled between t=0 and t=3. We can integrate the given velocity function to arrive at the position function. However, the distance traveled is irrespective of direction, so we must remember to take the absolute value of the velocity first.

The total distance traveled can be found using a definite integral. Namely, \int _{ 0 }^{ 3 }{ \left| sin({ t }^{ 2 }) \right| } dt=1.702

Example 5: (2007 AP Calculus AB Form B Exam Question 2 part a)

Image 2: The CollegeBoard Advanced Placement Program

We are asked for the total water entering the tank between t=0 and t=7. The rate of water entering the tank is f(t) and the total amount of water entering the tank can be found using a definite integral:

\int _{ 0 }^{ 7 }{ f(t) } dt=8264 gallons.

The Fundamental Theorem of Calculus

As you can see, the fundamental theorem of calculus establishes a procedure for calculating a definite integral. Now, this theorem on its own is already useful, but it also supplies us with the fact that this definite integral is equivalent to the total change over a particular interval, which comes in handy in a number of situations as we saw in the last two problems above.

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The Fundamental Theorem of Calculus (FTC) is one of the most important mathematical discoveries in history. You might think I’m exaggerating, but the FTC ranks up there with the Pythagorean Theorem and the invention of the numeral 0 in its elegance and wide-ranging applicability. In this article I will explain what the Fundamental Theorem of Calculus is and show how it is used.

What is the Fundamental Theorem of Calculus?

Although the main ideas were floating around beforehand, it wasn’t until the 1600s that Newton and Leibniz independently formalized calculus — including the Fundamental Theorem of Calculus.

Gottfried Wilhelm Leibniz (left) and Sir Isaac Newton (right). (Image of Leibniz courtesy of Ad Meskens)

The FTC establishes a direct link between two different branches of mathematics: analysis and geometry. Analysis deals with properties of functions and rates of change, while geometry can be used to measure things about shapes.

Informally, the FTC states that the area under a curve y = f(x) (a geometric measurement) can be found using an antiderivative of the function (an analytic tool). This correspondence between area and antiderivatives is what makes the theorem so important and useful.

Statement of the Theorem (Definite Integral Form)

There are actually two different forms of the Fundamental Theorem of Calculus. Here is the one that is used most often.

If f is continuous on a closed interval [a, b], then

In the above formula, F is antiderivative for f. That is, F is a function whose derivative is equal to f. Using mathematical notation, we would say that F ‘(x) = f(x).

Using the FTC

The Fundamental Theorem of Calculus provides a powerful tool for evaluating definite integrals. Here are the steps:

  1. Find an antiderivative for the integrand, using appropriate integration formulas.
  2. Plug the upper limit (b) and lower limit (a) of integration into the antiderivative F.
  3. Subtract to find the final answer: F(b) – F(a).

Example

In the following example, we work out a definite integral using the FTC. Remember, cos x is the derivative of sin x. So sin x is the antiderivative of cos x.

Applications of the FTC

Any time a definite integral needs to be evaluated, the Fundamental Theorem of Calculus can come to the rescue. One of the most common applications you’ll see on the AP Calculus exams is area under a curve.

Area Under a Curve

If a function f(x) is nonnegative on an interval [a, b], then the area of the region under the curve can be computed by a definite integral. The limits of integration, a and b, specify the left and right boundaries of the region. The bottom boundary is the x-axis, and the top boundary is the graph of f(x) itself.

Let’s see how it works in one of the simplest cases, f(x) = x. The graph is a diagonal line through the origin.

Suppose b > 0. We will find the area under y = f(x) = x between x = 0 and x = b.

First set up the definite integral that computes the area. Then, according to the Fundamental Theorem of Calculus, we just need to find an antiderivative for f(x) = x. You can verify that F(x) = x2/2 does the trick.

Areas and Antiderivatives

Now let’s try this same example (area under f(x) = x on [0, b]) but in a different way. By working the problem out using a different method, I hope to show you the remarkable connection between areas and antiderivatives.

Notice that the region under this particular function is simply a right triangle. You should know how to find the area of any triangle, using Area = (1/2)×(base)×(height). In this case, the base and height are the same: both are equal to b.

Therefore, by geometry we get Area = (1/2)×b×b = b2/2. The same result as we got above!

The fact that we get the same answer in this case might not be too surprising. However, the real power of the Fundamental Theorem of Calculus is that this link between areas and antiderivatives is true every single time. No matter how complicated the function is, you can find the area under the curve just using calculus.

The FTC and Accumulation Functions

There is a second part to the Fundamental Theorem of Calculus. It involves so-called accumulation functions. These are functions defined by a definite integral in which the upper limit of integration is the variable.

What is an Accumulation Function?

It’s helpful to think of an accumulation function as an “area so far” function. For any input x, the value of F(x) is the area under f from a to x. As x increases, more of the area gets “painted.”

For example, the accumulation function for f(x) = x with left endpoint a = 0 is F(x) = x2/2. This is because the area under y = x on the interval [0, x] is equal to the area of the triangle with base and height both each to x.

The Second Fundamental Theorem of Calculus

The second part of the FTC states that the accumulation function is just a particular antiderivative of the original function. Equivalently, the derivative of an accumulation function for a function f is equal to f(x) itself.

Here’s the formal statement of the theorem.

If f is continuous on a closed interval [a, b], then

Using the Second Fundamental Theorem

On the AP Calculus exam, you may be asked to take the derivative of an accumulation function. The FTC is just the right tool for the job. These problems can range from very easy to much more challenging, so let’s see a couple easy examples first.

Example Problem 1

In this problem, the FTC can be applied directly. Notice how the variable t gets swapped out and becomes x.

Example Problem 2

This time, we need to reverse the order of integration first. Remember, this introduces a negative.

The Chain Rule and the FTC

In the above examples, we could use the theorem because one of the limits of integration happened to be a single variable x. What if one or both of the limits of integration are functions of x? This situation can be handled with a more general formula that involves the Chain Rule.

Suppose u and v are both differentiable functions of x. Then

Here’s how it works:

Example Problem 3

Solution:

Final Thoughts

As with any topic on the AP Calculus exams, there is a lot to understand about the Fundamental Theorem of Calculus. However I hope that after having read this article, you have a much better understanding.

About Shaun Ault

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!


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