Now we turn our attention to the concept of the Definite Integral. The Definite Integral of a function is closely related to the antiderivative and indefinite integral of the function.the primary difference is that the definite integral, if it exists, is a real number value, while the other two represent an infinite number of functions that differ only by a constant.

The problem we are trying to solve using definite integrals is to find the area under the curve of a continuous function from x=a to x=b, above the x axis:

To set the stage for a powerful technique for finding the desired area, we find it is advantageous to allow the right-hand endpoint of the interval to be a variable. So the right-hand endpoint is x, a<x<b. Now, the area is a function of x, which we will call A(x), the shaded portion of the graph below.

Shaded area denotes A(x)

Let f(x)=3, on the interval [0,10]. We want to find the area of the region under the curve y=f(x)=3, over the interval [0,x].

y=f(x)=3

The area we are after is a rectangle with base width=x and height=3, so the area is: A(x)=3x. (Notice that 3x is an antiderivative of 3!)

Now consider the function f(x)=x-2, on the interval [2,8].

f(x)=x-2

Notice we want to find the area of a triangle with base length x-2 (because x-2 is the distance from 2 to x) and altitude x-2 (because f(x)=x-2), over the interval [2,x]. So A(x)=1/2*altitude*base = 1/2(x-2)(x-2).

Notice again that the answer is an antiderivative of the original function. Finding the areas under the curves on the given intervals was easy in the above examples because they are familiar geometric shapes with area formuas that are readily available. In most cases it is not so simple, as in the curve at the top of this page.

The first and most important step in solving the area problem for more general functions is to formally establish a connection between finding an area under a curve on an interval and antidifferentiation. That relationship is one of the cornerstones of calculus. The Fundamental Theorem has two parts. Text books and calculus professors go into this stuff in varrying degrees of depth, so I'll state what it is saying as simply as possible, and include the Symbolic statement for your referrence.

The first part says that the area under a curve on a given interval is an antiderivative of f(x):

Let f be continuous on an open interval I containing the number a and let for each x in the interval I. Then A is differentiable on I, and That is:

The use of the variable, t, is just fhe independent variable for f. We do this because we are using the letter x to denote the independent variable for y=A(x).

The second part says that we can find the area under the curve's actual value by evaluating the value of the function's antiderivative at b and subtracting the antiderivative's value at a.

The area we are looking for

The result is that the area under the curve on[1,3]=20/3. Notice that the constant, C, was not included in our calculation. It is not necessary because in the operation F(b)-F(a), you get C - C. So it holds that the numerical value of the definite integral of a function can be determined using ANY antiderivative of the function.

In the following section, we will develop an important alternate definition of the definite integral, the Riemann Integral. It's derrivation will confirm our methods for finding the value of a definite integral.