Module 2: Introduction

The goal of this module is to review the basic concepts about indefinite and definite integrals.

In this module you will complete the following activities:

Review the concept of antiderivative, definite and indefinite integrals.
Practice the covered concepts through some exercises, called "Now You Try".
Complete a quiz on this topic at the end of the module.

Suppose a rocket lifts off the surface of Earth with a constant acceleration of $20 m / s e c^{2}$ ${20 \; m/sec^2}$ We would like to know how fast will the rocket be going 1 minute later.

How fast the rocket is going 1 minute later is a different way of asking what is the velocity of the rocket at 1 minute. That is, if $v (t)$ $v(t)$ is the velocity of the rocket at time t (seconds), the question asks for $v (60)$ $v(60)$ The acceleration of the rocket is the instantaneous rate of change of the velocity, i.e., the derivative of the velocity. So we have

$\frac{d v}{d t} = 20 m / s e c^{2} .$ $\frac{dv}{dt} =20\; \; m/sec^2.$

We are looking for a function $v (t)$ $v(t)$ whose derivative is $20$ $20$ and $v (0) = 0$ $v(0) = 0$ (the rocket starts its journey with velocity 0). That is we are looking for the antiderivative of $20$ $20$ such that $v (0) = 0$ $v(0) = 0$ . Any function of the form $20 t + C$ $20t+C$ has a derivative of $20$ $20$ with respect to $t$ $t$ , where $c$ $c$ is any constant (number). But we want $20 (0) + C = 0$ $20(0)+C=0$ , therefore $C = 0 $ .$ $C=0$.$

The velocity function is

$v (t) = 20 t$ $v(t)=20t\$

and

$v (60) = 1200 m/sec \sim 2600 miles/hour$ $v(60)=1200\; \mbox{m/sec} \sim 2600 \; \mbox{miles/hour}$

$v (t) = 20 t$ $v(t)=20t$ is called an antiderivative of the function $20$ $20$ .