Fundamental Theorem of Calculus (FTC)

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

Expected Educational Results

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.

Bloom’s Taxonomy for different levels of understanding
Figure 1.1: Bloom's Taxonomy

Application of FTC

Definition: Velocity

Let s(t) be the displacement of some object over some values of time, t, then the velocity at time t is v(t)=s(t).

Definition: Acceleration

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

Definition: Displacement

Let v(t) be the velocity of some object from t=a to t=b, then the displacement (i.e., Net Change of Position) over the time interval [a,b] (by FTC) is s(t)=abv(t)dt.

Definition: Total Distance Traveled

Let v(t) be the velocity of some object from t=a to t=b, then the total distance traveled over the time interval [a,b] is total distance traveled=ab|v(t)|dt.

Area Under the Curve

Example 01:

Let v(t)=62t be the velocity (in m/s) of some object from t=0s to t=4s. Find the displacement over the time interval [0,4].

Solution: s(t)=04(62t)dt=(6tt2)|t=0t=4=(6(4)(4)2)(6(0)(0)2)=8

The object is 8m 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 Ablue=12bh=12(3)(6)=9 and the area of the red triangle is Ared=12bh=12(1)(2)=1. The Net Change Theorem suggests we can “add” these areas. However, since the red triangle is below the x-axis, we will subtract the area of the red triangle. Thus, the displacement of the object is s=AblueAred=91=8.

Example 02:

Let v(t)=62t be the velocity (in m/s) of some object from t=0s to t=4s. Find the total distance traveled over the time interval [0,4].

Solution:

s(t)=04|62t|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 Ablue=12bh=12(3)(6)=9 and the area of the red triangle is Ared=12bh=12(1)(2)=1. The Net Change Theorem suggests we can “add” these areas. Applying the absolute value to v(t) makes any negative y-value into a positive y-value. Since the red triangle in v(t) was below the x-axis, the red triangle in |v(t)| is above the x-axis. Thus, the total distance traveled of the object is total distance traveled=Ablue+Ared=9+1=10.

The object traveled a total distance of 10m.

Investigation 21

  1. Suppose an object moves along a straight line with velocity v(t)=6t3 (in meters/second)

    • Find the total distance traveled on 0t2

    • Find the displacement on 0t2

  2. Suppose an object moves along a straight line with acceleration a(t)=3t+4 (in m/s2), v(0)=3, 0t6

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

    • Find the total distance traveled

Homework

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

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 04:32 EDT Last Modified: Monday, 23 August 2021 - 06:06 (EDT)