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.

Application of FTC

Definition: Velocity

Let $s(t)$$s(t)$ be the displacement of some object over some values of time, $t$$t$, then the velocity at time $t$$t$ is v(t$)=s^{\prime }(t)$$)=s^{\,\prime}(t)$.

Definition: Acceleration

Let $s(t)$$s(t)$ be the displacement of some object over some values of time, $t$$t$, then the acceleration at time $t$$t$ is $a(t)=s^{\prime \prime }(t)$$a(t)=s^{\,\prime\prime}(t)$. Also, $a(t)=v^{\prime }(t)$$a(t)=v^{\,\prime}(t)$.

Definition: Displacement

Let $v(t)$$v(t)$ be the velocity of some object from $t=a$$t=a$ to $t=b$$t=b$, then the displacement (i.e., Net Change of Position) over the time interval $[a,b]$$[a,b]$ (by FTC) is $s(t)={\int }_a^{b}v(t)dt$$s(t)=\int_a^b{v(t)\,dt}$.

Definition: Total Distance Traveled

Let $v(t)$$v(t)$ be the velocity of some object from $t=a$$t=a$ to $t=b$$t=b$, then the total distance traveled over the time interval $[a,b]$$[a,b]$ is $\text{total distance traveled}={\int }_a^b|v(t)|dt$$\text{total distance traveled}=\int_a^b{|v(t)|\,dt}$.

Area Under the Curve

Example 01:

Let $v(t)=6-2t$$v(t)=6-2t$ be the velocity (in $m$$m$/$s$$s$) of some object from $t=0s$$t=0s$ to $t=4s$$t=4s$. Find the displacement over the time interval $[0,4]$$[0,4]$.

Solution: $s(t)={\int }_0^4(6-2t)dt={(6t-t^2)|}_{t=0}^{t=4}=(6(4)-(4{)}^2)-(6(0)-(0{)}^2)=8$$s(t)=\int_0^4{\left(6-2t\right)\,dt} = \left.\left(6t-t^2\right)\right|_{t=0}^{t=4} = \left(6(4)-(4)^2\right)-\left(6(0)-(0)^2\right) = 8$

The object is $8m$$8 m$ from the initial starting point.

The displacement is the net area under the curve. This is shown in the Figure below:

NOTE: The area of the blue triangle is ${\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9}$$\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9$ and the area of the red triangle is ${\displaystyle A_{red}=\frac{1}{2}bh=\frac{1}{2}(1)(2)=1}$$\displaystyle A_{red}=\frac{1}{2}bh=\frac{1}{2}(1)(2)=1$. The Net Change Theorem suggests we can “add” these areas. However, since the red triangle is below the $x$$x$-axis, we will subtract the area of the red triangle. Thus, the displacement of the object is $s=A_{blue}-A_{red}=9-1=8$$s=A_{blue}-A_{red}=9-1=8$.

Example 02:

Let $v(t)=6-2t$$v(t)=6-2t$ be the velocity (in $m$$m$/$s$$s$) of some object from $t=0s$$t=0s$ to $t=4s$$t=4s$. Find the total distance traveled over the time interval $[0,4]$$[0,4]$.

Solution:

$s(t)={\int }_0^4|6-2t|dt$$s(t)=\int_0^4{\left|6-2t\right|\,dt}$

The total distance traveled is the net area under the absolute value of the curve. This is shown in the Figure below:

NOTE: The area of the blue triangle is ${\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9}$$\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9$ and the area of the red triangle is ${\displaystyle A_{red}=\frac{1}{2}bh=\frac{1}{2}(1)(2)=1}$$\displaystyle A_{red}=\frac{1}{2}bh=\frac{1}{2}(1)(2)=1$. The Net Change Theorem suggests we can “add” these areas. Applying the absolute value to $v(t)$$v(t)$ makes any negative $y$$y$-value into a positive $y$$y$-value. Since the red triangle in $v(t)$$v(t)$ was below the $x$$x$-axis, the red triangle in $|v(t)|$$|v(t)|$ is above the $x$$x$-axis. Thus, the total distance traveled of the object is $\text{total distance traveled}=A_{blue}+A_{red}=9+1=10$$\text{total distance traveled}=A_{blue}+A_{red}=9+1=10$.

The object traveled a total distance of $10m$$10 m$.

Investigation 21

1. Suppose an object moves along a straight line with velocity $v(t)=6t-3$$v(t)=6t-3$ (in meters/second)

   • Find the total distance traveled on $0\le t\le 2$$0\leq t\leq 2$

   • Find the displacement on $0\le t\le 2$$0\leq t\leq 2$

2. Suppose an object moves along a straight line with acceleration $a(t)=3t+4$$a(t)=3t+4$ (in $\text{m}/{\text{s}}^2$$\text{m}/\text{s}^2$), $v(0)=3$$v(0)=3$, $0\le t\le 6$$0\leq t\leq 6$

   • Find the velocity at time $t$$t$, $v(t)$$v(t)$

   • Find the total distance traveled

Homework

At this time, you should be able to complete the following assignments:

• Section 5.4: # 69, 71.

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Created: Tuesday, 25 August 2020 04:32 EDT Last Modified: Monday, 23 August 2021 - 06:06 (EDT)