Chapter 4 Integration

4.1 Antidifferentiation and Indefinate Integrals

The work to this point has involved finding and applying the first or second derivative of a function. In the this chapter we will reverse the process. If we know the derivative of a function how do we obtain the original function? The process is called antidifferentiation or integration.

Definition of an Antiderivative

For example to find the antiderivative of the function f(x)=2x we need to find the function that has derivative of 2x. What function has a derivative of 2x? How many functions have a derivative of 2x?

So, has derivative 2x we say that is an antiderivative of 2x. In fact + or - any constant is an antiderivative of 2x.

The symbol used to indicate this process is and is called an integral sign. The notation is read the indefinite integral of f(x) and denotes the family of antiderivatives of f(x). f(x) is called the integrand and dx is the operator.

F(x) + c denotes the family of antiderivatives of f(x) and c is called the constant of integration. Since it follows that

The rules for antidifferentiation are the steps that reverse the steps for differentiation. Each rule for finding a derivative has its corresponding rule of integration.

Power rule: to integrate a power of x, add one to the exponent then divide by the new exponent.

Integrate

Constant rule: the antiderivative of a constant is the constant times x

Other rules of indefinite integrals:

Find each antiderivative

Be sure to take the anti- derivative of each term in the same step!

Example : Evaluate

Theorem 4.1 Representation of Antiderivatives

Initial Conditions and Particular Solutions When we find antiderivatives and add the constant C, we are creating a family of curves for each value of C. C=0 C=1 C=3

Basic Integration Rules

A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. And An initial acceleration of -32 feet per second. A. Find the position function giving the height S as a function of the time t. B. When does the ball hit the ground? Begin by integrating, we get The initial velocity is 64 feet per second. This means at t = 0, S’ = 64 Integrating again, we can find, s(t).

C = 80 Thus, the position function is: B. When will the ball hit the ground? When S = 0 Since t must be positive, the ball hit the ground after 5 seconds. t = 5, t = -1 The initial height means at t = 0, S = 80

2.Find a general solution and show a rough graph for y if 3.Find an equation for y if and y passes through (1,1).

2.Find a general solution and show a rough graph for y if 3.Find an equation for y if and y passes through (1,1).

Solve the differential equation