# 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 CalculusPart I.

• Objective 01–03: I can explain the meaning of the Fundamental Theorem of CalculusPart 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 CalculusPart I.

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

• Objective 01–07: I can use the Net Change Theorem to identify the meaning of ${\int }_{a}^{b}{f}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(x\right)\phantom{\rule{0.167em}{0ex}}\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\left(t\right)$$s(t)$ be the displacement of some object over some values of time, $t$$t$, then the velocity at time $t$$t$ is v(t$\right)={s}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(t\right)$$)=s^{\,\prime}(t)$.

### Definition: Acceleration

Let $s\left(t\right)$$s(t)$ be the displacement of some object over some values of time, $t$$t$, then the acceleration at time $t$$t$ is $a\left(t\right)={s}^{\phantom{\rule{0.167em}{0ex}}\prime \prime }\left(t\right)$$a(t)=s^{\,\prime\prime}(t)$. Also, $a\left(t\right)={v}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(t\right)$$a(t)=v^{\,\prime}(t)$.

### Definition: Displacement

Let $v\left(t\right)$$v(t)$ be the velocity of some object from $t=a$$t=a$ to $t=b$$t=b$, then the displacement (i.e., Net Change of Position) over the time interval $\left[a,b\right]$$[a,b]$ (by FTC) is $s\left(t\right)={\int }_{a}^{b}v\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$$s(t)=\int_a^b{v(t)\,dt}$.

### Definition: Total Distance Traveled

Let $v\left(t\right)$$v(t)$ be the velocity of some object from $t=a$$t=a$ to $t=b$$t=b$, then the total distance traveled over the time interval $\left[a,b\right]$$[a,b]$ is $\text{total distance traveled}={\int }_{a}^{b}|v\left(t\right)|\phantom{\rule{0.167em}{0ex}}dt$$\text{total distance traveled}=\int_a^b{|v(t)|\,dt}$.

### Area Under the Curve

Example 01:

Let $v\left(t\right)=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$$t=4s$. Find the displacement over the time interval $\left[0,4\right]$$[0,4]$.

Solution: $s\left(t\right)={\int }_{0}^{4}\left(6-2t\right)\phantom{\rule{0.167em}{0ex}}dt={\left(6t-{t}^{2}\right)|}_{t=0}^{t=4}=\left(6\left(4\right)-\left(4{\right)}^{2}\right)-\left(6\left(0\right)-\left(0{\right)}^{2}\right)=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 ${A}_{blue}=\frac{1}{2}bh=\frac{1}{2}\left(3\right)\left(6\right)=9$$\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9$ and the area of the red triangle is ${A}_{red}=\frac{1}{2}bh=\frac{1}{2}\left(1\right)\left(2\right)=1$$\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\left(t\right)=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$$t=4s$. Find the total distance traveled over the time interval $\left[0,4\right]$$[0,4]$.

Solution:

$s\left(t\right)={\int }_{0}^{4}|6-2t|\phantom{\rule{0.167em}{0ex}}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 ${A}_{blue}=\frac{1}{2}bh=\frac{1}{2}\left(3\right)\left(6\right)=9$$\displaystyle A_{blue}=\frac{1}{2}bh=\frac{1}{2}(3)(6)=9$ and the area of the red triangle is ${A}_{red}=\frac{1}{2}bh=\frac{1}{2}\left(1\right)\left(2\right)=1$$\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\left(t\right)$$v(t)$ makes any negative $y$$y$-value into a positive $y$$y$-value. Since the red triangle in $v\left(t\right)$$v(t)$ was below the $x$$x$-axis, the red triangle in $|v\left(t\right)|$$|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\left(t\right)=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\left(t\right)=3t+4$$a(t)=3t+4$ (in $\text{m}/{\text{s}}^{2}$$\text{m}/\text{s}^2$), $v\left(0\right)=3$$v(0)=3$, $0\le t\le 6$$0\leq t\leq 6$

• Find the velocity at time $t$$t$, $v\left(t\right)$$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.