Fundamental Theorem of Calculus (FTC)

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 4.9 – Antiderivatives

• Section 5.3 – The Fundamental Theorem of Calculus

• Section 5.4 – Indefinite Integrals and the Net Change Theorem

Expected Educational Results

• Objective 01–01: I can state the Fundamental Theorem of Calculus.

• Objective 01–02: I can explain the meaning of the Fundamental Theorem of Calculus – Part I.

• Objective 01–03: I can explain the meaning of the Fundamental Theorem of Calculus – Part II.

• Objective 01–04: I can determine the properties of an area function using FTC-I and Calculus I.

• Objective 01–05: I can evaluate indefinite integrals using the Fundamental Theorem of Calculus – Part I.

• Objective 01–06: I can evaluate definite integrals using the Fundamental Theorem of Calculus – Part II.

• Objective 01–07: I can use the Net Change Theorem to identify the meaning of ${\int }_a^b{f}^{\prime }(x)\text{d}x$$\int_a^b{f^{\,\prime}(x)\,\text{d}x}$.

• Objective 01–08: Given velocity of an object in motion along a line, I can find the object’s displacement and the total distance traveled.

• Objective 01–09: Given acceleration of an object in motion along a line, I can find the object’s velocity, displacement, and the total distance traveled.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Fundamental Theorem of Calculus – Part I (FTC-I)

Find Derivatives of Integral Functions

Example: Find ${\displaystyle \frac{d}{dx}{\int }_{-2}^x(3t^2+5t)\text{d}t}$$\displaystyle\frac{d}{dx}\int_{-2}^x{(3t^2+5t)\,\text{d}t}$.

Solution – Method 01:

NOTE: This method needs more information that we currently have discussed in this course; however, you discussed the ideas needed for this method in Math 2211 – Calculus of a Single Variable – I.

There are three analytic steps to perform:

1. Find the antiderivative: ${\displaystyle F(t)=\int (3t^2+5t)\text{d}t}$$\displaystyle F(t)=\int{(3t^2+5t)\,\text{d}t}$

   • In this Example, use the Power Rule for Antiderivatives: $F(t)=t^3+\frac{5}{2}{t}^2+C$$F(t)=t^3+\frac{5}{2}t^2+C$

   • NOTE: Not all functions, $f(x)$$f(x)$, have analytic (primitive) antiderivatives.

2. Evaluate ${\int }_{-2}^x(3t^2+5t)\text{d}t=F(x)-F(-2)$$\int_{-2}^x{(3t^2+5t)\,\text{d}t}={\color{red}F(x)}-{\color{blue}F(-2)}$

   • $x^3+\frac{5}{2}{x}^2+C-\left(t^3+\frac{5}{2}{t}^2+C\right)$${\color{red}x^3+\frac{5}{2}x^2+C}-\left({\color{blue}t^3+\frac{5}{2}t^2+C}\right)$

3. Differentiate with respect to $\mathbf{x}$$\mathbf{x}$ the result in Step 2:

   • $\frac{d}{dx}[x^3+\frac{5}{2}{x}^2+C-\left(t^3+\frac{5}{2}{t}^2+C\right)]$$\frac{d}{dx}\left[{\color{red}x^3+\frac{5}{2}x^2+C}-\left({\color{blue}t^3+\frac{5}{2}t^2+C}\right)\right]$ $\Rightarrow \frac{d}{dx}{x}^3+\frac{d}{dx}\frac{5}{2}{x}^2+\frac{d}{dx}C-\frac{d}{dx}\left(t^3+\frac{5}{2}{t}^2+C\right)$$\Rightarrow\frac{d}{dx}{\color{red}x^3}+\frac{d}{dx}{\color{red}\frac{5}{2}x^2}+\frac{d}{dx}{\color{red}C}-\frac{d}{dx}\left({\color{blue}t^3+\frac{5}{2}t^2+C}\right)$ $\Rightarrow 3x^2+5x+0-0=3x^2+5x$$\Rightarrow {\color{red}3x^2}+{\color{red}5x}+{\color{red}0}-{\color{blue}0}=3x^2+5x$

4. Compare the answer in Step 3 to the integrand in the question.

Solution – Method 02:

Verify the conditions:

• $3t^2+5t$$3t^2+5t$ is continuous on the interval $[-2,\mathrm{\infty })$$[-2,\infty)$

• The lower limit of integration is a constant, $a=-2$$a=-2$

• The upper limit of integration is $x$$x$

Then, by the Fundamental Theorem of Calculus – Part I (FTC-I), ${\displaystyle \frac{d}{dx}{\int }_{-2}^x(3t^2+5t)\text{d}t=3x^2+5x}$$\displaystyle\frac{d}{dx}\int_{-2}^x{(3t^2+5t)\,\text{d}t}=3x^2+5x$

NOTE: Method 02 has little to no computations, so Method 02 is the desired method to use and is the expected method for an Assessment. FTC-I is always valid (even for functions that do not have analytic antiderivatives) if the conditions of the Theorem are met.

Activity 08

NOTE: This Mathematica activity will help you quickly verify and understand the Fundamental Theorem of Calculus - Part I. You will not need to use the following code on any Assessment.

Let $g(x)={\int }_{-2}^x(3t^2+5t)dt$$g(x)=\int_{-2}^x{\left(3t^2+5t\right)\,dt}$. Use Mathematica to find $g^{\prime }(x)=\frac{d}{dx}{\int }_{-2}^x(3t^2+5t)dt$$g^{\,\prime}(x)=\frac{d}{dx}\int_{-2}^x{\left(3t^2+5t\right)\,dt}$.

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(* Find g'(x) *)
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D[Integrate[3t^2+5t,{t,-2,x}],x]

Warnings:

1. You do not need to change the above code in any manner for this Activity. This code works in Mathematica 13.0 and above.

2. This code has two nested Mathematica commands:

   • The Integrate[ ] command (in red): $\text{D[}\text{Integrate[3t}\wedge \text{2+5t,{t,-2,x}]}\text{,x]}$$\text{{\color{blue}D[}\color{red}Integrate[3t}{\color{red}\wedge}\text{{\color{red}2+5t,\{t,-2,x\}]}\color{blue},x]}$

     • Integrate[ ] has at least two arguments:

       • The integrand function: $\text{3t}\wedge \text{2+5t}$$\text{3t}\!\wedge\!\text{2+5t}$

       • The independent variable and the limits of integration, grouped with braces $\{$$\{$ and $\}$$\}$: $\{t,-2,x\}$$\{t,-2,x\}$

     • By FTC-I, this integral is an area function and is a function of $x$$x$.

   • The derivative command, D[ ], (in blue):

     • D[ ] has at least two arguments:

       • The function, in this case, the area function (in red)

       • The independent variable of the area function: $x$$x$

3. For help on using the Integrate[ ] function:

   1. In Mathematica, execute the code: $\text{?Integrate}$$\text{?Integrate}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the Integrate[ ] function – you will be able to copy-paste code.

4. For help on using the D[ ] function:

   1. In Mathematica, execute the code: $\text{?D}$$\text{?D}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the D[ ] function – you will be able to copy-paste code.

5. Remember, correct Mathematica code will be all black except for variables.

6. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Activity 09

NOTE: This Mathematica activity will help you quickly verify and understand the Fundamental Theorem of Calculus - Part I. You will not need to use the following code on any Assessment.

Let ${\displaystyle g(x)={\int }_x^{-2}(3t^2+5t)dt}$$\displaystyle g(x)=\int_x^{-2}{\left(3t^2+5t\right)\,dt}$. Use Mathematica to find $g^{\prime }(x)=\frac{d}{dx}{\int }_x^{-2}(3t^2+5t)dt$$g^{\,\prime}(x)=\frac{d}{dx}\int_x^{-2}{\left(3t^2+5t\right)\,dt}$.

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1
(* Find g'(x) *)
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D[Integrate[3t^2+5t,{t,x,-2}],x]

1. What do you notice about the limits of integration in Activity 08 compared to the limits of integration in Activity 09? Explain.

2. How is the answer to Activity 09 different from the Activity 08? Use one of the Properties of Definite Integrals (see PRE_01v_FTC_Definite_Integrals_Props.html) to explain this result.

Activity 10

Let ${\displaystyle g(x)={\int }_{-2}^{x^2}(3t^2+5t)dt}$$\displaystyle g(x)=\int_{-2}^{x^2}{\left(3t^2+5t\right)\,dt}$. Use Mathematica to find $g^{\prime }(x)=\frac{d}{dx}{\int }_{-2}^{x^2}(3t^2+5t)dt$$g^{\,\prime}(x)=\frac{d}{dx}\int_{-2}^{x^2}{\left(3t^2+5t\right)\,dt}$.

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2
1
(* Find g'(x) *)
2
D[Integrate[3t^2+5t,{t,-2,x^2}],x]

1. Verify the result from the above code is $6x^5+10x^3$$6x^5+10x^3$.

2. Verify the result can be rewritten as: $2x(3{\left(x^2\right)}^2+5\left(x^2\right))$${\color{blue}2x}\left(3\left({\color{red}x^2}\right)^2+5\left({\color{red}x^2}\right)\right)$

3. How is the answer to Activity 10 similar to the answer from Activity 08? Explain.

4. How is the answer to Activity 10 different from the answer from Activity 08? Explain.

5. What is the derivative of the “inside” function, $x^2$${\color{red}x^2}$? What differentiation rule is needed to find $g^{\prime }(x)$$g^{\,\prime}(x)$? Explain.

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Tuesday, 25 August 2020 03:34 EDT Last Modified: Monday, 30 May 2022 - 13:25 (EDT)