# 6.4 Fundamental Theorem of Calculus

## Learning Objectives

- Describe the meaning of the Mean Value Theorem for Integrals.
- State the meaning of the Fundamental Theorem of Calculus, Part 1.
- Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals.
- State the meaning of the Fundamental Theorem of Calculus, Part 2.
- Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals.
- Explain the relationship between differentiation and integration.

In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some more powerful and useful techniques for evaluating definite integrals.

These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the **Fundamental Theorem of Calculus**, which has two parts that we examine in this section. Its very name indicates how central this theorem is to the entire development of calculus.

Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, Britannica’s page on Newton with multimedia clips.

Before we get to this crucial theorem, however, let’s examine another important theorem, the Mean Value Theorem for Integrals, which is needed to prove the Fundamental Theorem of Calculus.

## The Mean Value Theorem for Integrals

The **Mean Value Theorem for Integrals** states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if [latex]f(x)[/latex] is continuous, a point [latex]c[/latex] exists in an interval [latex]\left[a,b\right][/latex] such that the value of the function at [latex]c[/latex] is equal to the average value of [latex]f(x)[/latex] over [latex]\left[a,b\right].[/latex] We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

### The Mean Value Theorem for Integrals

If [latex]f(x)[/latex] is continuous over an interval [latex]\left[a,b\right],[/latex] then there is at least one point [latex]c\in \left[a,b\right][/latex] such that

$$f(c)=\frac{1}{b-a}{\int }_{a}^{b}f(x)dx.$$

This formula can also be stated as

$${\int }_{a}^{b}f(x)dx=f(c)(b-a).$$

## Proof

Since [latex]f(x)[/latex] is continuous on [latex]\left[a,b\right],[/latex] by the extreme value theorem (see Maxima and Minima), it assumes minimum and maximum values—[latex]m[/latex] and *M*, respectively—on [latex]\left[a,b\right].[/latex] Then, for all [latex]x[/latex] in [latex]\left[a,b\right],[/latex] we have [latex]m\le f(x)\le M.[/latex] Therefore, by the comparison theorem (see The Definite Integral), we have

$$m(b-a)\le {\int }_{a}^{b}f(x)dx\le M(b-a).$$

Dividing by [latex]b-a[/latex] gives us

$$m\le \frac{1}{b-a}{\int }_{a}^{b}f(x)dx\le M.$$

Since [latex]\frac{1}{b-a}{\int }_{a}^{b}f(x)dx[/latex] is a number between [latex]m[/latex] and *M*, and since [latex]f(x)[/latex] is continuous and assumes the values [latex]m[/latex] and *M* over [latex]\left[a,b\right],[/latex] by the Intermediate Value Theorem (see Continuity), there is a number [latex]c[/latex] over [latex]\left[a,b\right][/latex] such that

$$f(c)=\frac{1}{b-a}{\int }_{a}^{b}f(x)dx,$$

and the proof is complete.

## Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the **Fundamental Theorem of Calculus, Part 1**, is stated here. Part 1 establishes the relationship between differentiation and integration.

### Fundamental Theorem of Calculus, Part 1

If [latex]f(x)[/latex] is continuous over an interval [latex]\left[a,b\right],[/latex] and the function [latex]F(x)[/latex] is defined by

$$F(x)={\int }_{a}^{x}f(t)dt,$$

then [latex]{F}^{\prime }(x)=f(x)[/latex] over [latex]\left[a,b\right].[/latex]

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, [latex]F(x),[/latex] as the definite integral of another function, [latex]f(t),[/latex] from the point [latex]a[/latex] to the point [latex]x.[/latex]

At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of [latex]x[/latex], the definite integral is a number. So the function [latex]F(x)[/latex] returns a number (the value of the definite integral) for each value of [latex]x.[/latex]

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the *Fundamental* Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.

### Proof

Applying the definition of the derivative, we have

$$\begin{array}{}\\ \\ \\ {F}^{\prime }(x)\hfill & =\underset{h\to 0}{\text{lim}}\frac{F(x+h)-F(x)}{h}\hfill \\ \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{h}\left[{\int }_{a}^{x+h}f(t)dt-{\int }_{a}^{x}f(t)dt\right]\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{h}\left[{\int }_{a}^{x+h}f(t)dt+{\int }_{x}^{a}f(t)dt\right]\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{h}{\int }_{x}^{x+h}f(t)dt.\hfill \end{array}$$

Looking carefully at this last expression, we see [latex]\frac{1}{h}{\int }_{x}^{x+h}f(t)dt[/latex] is just the average value of the function [latex]f(x)[/latex] over the interval [latex]\left[x,x+h\right].[/latex] Therefore, by (Figure 6.15), there is some number [latex]c[/latex] in [latex]\left[x,x+h\right][/latex] such that

$$\frac{1}{h}{\int }_{x}^{x+h}f(x)dx=f(c).$$

In addition, since [latex]c[/latex] is between [latex]x[/latex] and [latex]h[/latex], [latex]c[/latex] approaches [latex]x[/latex] as [latex]h[/latex] approaches zero. Also, since [latex]f(x)[/latex] is continuous, we have [latex]\underset{h\to 0}{\text{lim}}f(c)=\underset{c\to x}{\text{lim}}f(c)=f(x).[/latex] Putting all these pieces together, we have

$$\begin{array}{}\\ {F}^{\prime }(x)\hfill & =\underset{h\to 0}{\text{lim}}\frac{1}{h}{\int }_{x}^{x+h}f(x)dx\hfill \\ & =\underset{h\to 0}{\text{lim}}f(c)\hfill \\ & =f(x),\hfill \end{array}$$

and the proof is complete.

## Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The **Fundamental Theorem of Calculus, Part 2**, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

After finding approximate areas by adding the areas of [latex]n[/latex] rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.

### The Fundamental Theorem of Calculus, Part 2

If [latex]f[/latex] is continuous over the interval [latex]\left[a,b\right][/latex] and [latex]F(x)[/latex] is any antiderivative of [latex]f(x),[/latex] then

$${\int }_{a}^{b}f(x)dx=F(b)-F(a).$$

We often see the notation [latex]{F(x)|}_{a}^{b}[/latex] to denote the expression [latex]F(b)-F(a).[/latex] We use this vertical bar and associated limits [latex]a[/latex] and [latex]b[/latex] to indicate that we should evaluate the function [latex]F(x)[/latex] at the upper limit (in this case, [latex]b[/latex]), and subtract the value of the function [latex]F(x)[/latex] evaluated at the lower limit (in this case, [latex]a[/latex]).

The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

## Proof

Let [latex]P=\left\{{x}_{i}\right\},i=0,1\text{,…,}n[/latex] be a regular partition of [latex]\left[a,b\right].[/latex] Then, we can write

[latex]F(b)-F(a) =F({x}_{n})-F({x}_{0})[/latex]

[latex]=\left[F({x}_{n})-F({x}_{n-1})\right]+\left[F({x}_{n-1})-F({x}_{n-2})\right]+\text{…}+\left[F({x}_{1})-F({x}_{0})\right][/latex]

[latex]=\underset{i=1}{\overset{n}{\text{∑}}}\left[F({x}_{i})-F({x}_{i-1})\right].[/latex]

Now, we know *F* is an antiderivative of [latex]f[/latex] over [latex]\left[a,b\right],[/latex] so by the Mean Value Theorem (see The Mean Value Theorem) for [latex]i=0,1\text{,…,}n[/latex] we can find [latex]{c}_{i}[/latex] in [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] such that

$$F({x}_{i})-F({x}_{i-1})={F}^{\prime }({c}_{i})({x}_{i}-{x}_{i-1})=f({c}_{i})\text{Δ}x.$$

Then, substituting into the previous equation, we have

$$F(b)-F(a)=\underset{i=1}{\overset{n}{\text{∑}}}f({c}_{i})\text{Δ}x.$$

Taking the limit of both sides as [latex]n\to \infty ,[/latex] we obtain

$$\begin{array}{}\\ \\ F(b)-F(a)\hfill & =\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}f({c}_{i})\text{Δ}x\hfill \\ & ={\int }_{a}^{b}f(x)dx.\hfill \end{array}$$

## Student Project

### A Parachutist in Free Fall

Julie is an avid skydiver. She has more than 300 jumps under her belt and has mastered the art of making adjustments to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the ground, she reaches a terminal velocity of approximately 120 mph (176 ft/sec). If, instead, she orients her body with her head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft/sec).

Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplify our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately starts falling at a velocity given by [latex]v(t)=32t.[/latex] She continues to accelerate according to this velocity function until she reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcord and slows down to land.

On her first jump of the day, Julie orients herself in the slower “belly down” position (terminal velocity is 176 ft/sec). Using this information, answer the following questions.

- How long after she exits the aircraft does Julie reach terminal velocity?
- Based on your answer to question 1, set up an expression involving one or more integrals that represents the distance Julie falls after 30 sec.
- If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?
- Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down, during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft/sec. Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch the ground.

On Julie’s second jump of the day, she decides she wants to fall a little faster and orients herself in the “head down” position. Her terminal velocity in this position is 220 ft/sec. Answer these questions based on this velocity: - How long does it take Julie to reach terminal velocity in this case?
- Before pulling her ripcord, Julie reorients her body in the “belly down” position so she is not moving quite as fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend in a free fall before beginning the reorientation?

Some jumpers wear “wingsuits” like in this figure. These suits have fabric panels between the arms and legs and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes called “flying squirrel suits.”) When wearing these suits, terminal velocity can be reduced to about 30 mph (44 ft/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast to land safely.

Answer the following question based on the velocity in a wingsuit.

- If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, how long does she get to spend gliding around in the air?