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.

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.

Watch the following video to see the worked solution to Example: Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration.

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.

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]\begin{array}{cc}F(b)-F(a)\hfill & =F({x}_{n})-F({x}_{0})\hfill \\ & =\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]\hfill \\ \\ & =\underset{i=1}{\overset{n}{\text{∑}}}\left[F({x}_{i})-F({x}_{i-1})\right].\hfill \end{array}[/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

[latex]F({x}_{i})-F({x}_{i-1})={F}^{\prime }({c}_{i})({x}_{i}-{x}_{i-1})=f({c}_{i})\text{Δ}x.[/latex]

 

Then, substituting into the previous equation, we have

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

 

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

[latex]\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 \\ & ={\displaystyle\int }_{a}^{b}f(x)dx.\hfill \end{array}[/latex]

[latex]_\blacksquare[/latex]

Example: Evaluating an Integral with the Fundamental Theorem of Calculus

Use the second part of the Fundamental Theorem of Calculus to evaluate

[latex]{\displaystyle\int }_{-2}^{2}({t}^{2}-4)dt.[/latex]

Show Solution

Recall the power rule for Antiderivatives:

[latex]\text{ If }y={x}^{n},\displaystyle\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C.[/latex]

Use this rule to find the antiderivative of the function and then apply the theorem. We have

[latex]\begin{array}{cc}{\displaystyle\int }_{-2}^{2}({t}^{2}-4)dt\hfill & =\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\hfill \\ \\ \\ & =\left[\frac{{(2)}^{3}}{3}-4(2)\right]-\left[\frac{{(-2)}^{3}}{3}-4(-2)\right]\hfill \\ & =(\frac{8}{3}-8)-(-\frac{8}{3}+8)\hfill \\ & =\frac{8}{3}-8+\frac{8}{3}-8\hfill \\ & =\frac{16}{3}-16\hfill \\ & =-\frac{32}{3}.\hfill \end{array}[/latex]

Analysis

Notice that we did not include the “+ C” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative with [latex]C=0.[/latex] If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.

The region of the area we just calculated is depicted in Figure 3. Note that the region between the curve and the [latex]x[/latex]-axis is all below the [latex]x[/latex]-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.

Figure 3. The evaluation of a definite integral can produce a negative value, even though area is always positive.

Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2

Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:

[latex]{\displaystyle\int }_{1}^{9}\dfrac{x-1}{\sqrt{x}}dx.[/latex]

Show Solution

First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:

[latex]{\displaystyle\int }_{1}^{9}\frac{x-1}{{x}^{1\text{/}2}}dx={\displaystyle\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx\text{.}[/latex]

Use the properties of exponents to simplify:

[latex]{\displaystyle\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx={\displaystyle\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx\text{.}[/latex]

Now, integrate using the power rule:

[latex]\begin{array}{}\\ \\ {\displaystyle\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx\hfill & ={(\frac{{x}^{3\text{/}2}}{\frac{3}{2}}-\frac{{x}^{1\text{/}2}}{\frac{1}{2}})|}_{1}^{9}\hfill \\ \\ & =\left[\frac{{(9)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(9)}^{1\text{/}2}}{\frac{1}{2}}\right]-\left[\frac{{(1)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(1)}^{1\text{/}2}}{\frac{1}{2}}\right]\hfill \\ & =\left[\frac{2}{3}(27)-2(3)\right]-\left[\frac{2}{3}(1)-2(1)\right]\hfill \\ & =18-6-\frac{2}{3}+2\hfill \\ & =\frac{40}{3}.\hfill \end{array}[/latex]

See Figure 4.

Figure 4. The area under the curve from [latex]x=1[/latex] to [latex]x=9[/latex] can be calculated by evaluating a definite integral.

Watch the following video to see the worked solution to Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2.

Watch the following video to see the worked solution to Example: A Roller-Skating Race.

Activity: A Parachutist in Free Fall

 

Figure 5. Skydivers can adjust the velocity of their dive by changing the position of their body during the free fall. (credit: Jeremy T. Lock)

 

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.

1. How long after she exits the aircraft does Julie reach terminal velocity?
2. 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.
3. If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?
4. 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:
5. How long does it take Julie to reach terminal velocity in this case?
6. 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” (see Figure 6). 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.
   

   Figure 6. The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a skydiver’s fall. (credit: Richard Schneider)

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

7. 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?