There are many calculus problems where the derivative of a function is known and the function is desired. For example, if a mathematical expression for the rate of population growth dP/dT is known, is it possible to "work backwards" to find the expression for P, how the population varies over time?

 The process of starting with a derivative and working back to the function is quite naturally called the antiderivative. The antiderivative of a function is an easy concept but often is operationally difficult. There are many integragtion problems where finding the antiderivative will prove a major challenge.

 In some problems the integral can be viewed as the area under the curve of the function being integrated. This is often very helpful in getting a physical "feel" for the problem and the process of integration. This view of the integral will be discussed later in the chapter.

 Some problems in integration require a great deal of imaginative thinking and manipulative ability. The simplest first approach to integration is via the antiderivative. After that we will move on to using the area under the curve approach and finally to the more difficult integral problems.

Start with a simple function, y = x². The derivative of that function is written as

dy = 2xdx and finally as dy/dx = 2x.

Keeping this short review of differentiation in mind, suppose we encountered a derivative

and want to know how u varies with v. Keep the differential (of y = x²) in front of you and just work backwards

du/dv = 2v can be written as du = 2vdv

Now all we need to do is perform the inverse or "anti" derivative operation to find u in terms of v. This being mathematics, no operation can be performed without a symbol. For integration we use this elongated "s" shape, so write

The left side of this equation is the integral of the differential, two inverse operations. The d acting on u is the derivative while the elongated "s" shape acting on du is the antiderivative. The result of these inverse operations on u is that the left side of this equation is u. The operation is somewhat like squaring a square root. The right side is not so easy except that we have the differential example just above us. The differential of x² is 2xdx so the integral of 2vdv is v². The function described by the differential statement du = 2vdv is therefore u = v².

 Conceptually the antiderivative is not difficult. Actually finding the antiderivative of a complicated function is often not at all easy. Polynomials are the easiest to work with and that is where we will start.