As explained in the chapter titled “Integration Basics”, the fundamental theorem of calculus tells us that to evaluate the area under a curve from to , we first evaluate the anti-derivative of
and then evaluate . That is, area under the curve from to is
Readers who have even the slightest doubt regarding the discussion above are advised to refer to the chapter on “Integration Basics” before reading on.
Definite integration is not all about just evaluating the anti-derivative and substituting the upper and lower limits. Working through this chapter, you will realise that a lot of techniques exist which help us in evaluating the definite integral without resorting to the (many times tedious) process of first determining the anti-derivative. We will develop all these techniques one by one from scratch, starting with some extremely basic properties in Section –
Suppose that on some interval . Then, the area under the curve from to will be negative in sign, i.e
This is obvious once you consider how the definite integral was arrived at in the first place; as a limit of the sum of the rectangles . Thus, if in some interval then the area of the rectangles in that interval will also be negative.
This property means that for example, if has the following form
then will equal and not .
If we need to evaluate (the magnitude of the bounded area), we will have to calculate
From this, it should also be obvious that
The area under the curve from to is equal in magnitude but opposite in sign to the area under the same curve from to , i.e
This property is obvious if you consider the Newton-Leibnitz formula. If is the anti-derivative of , then is while .
The area under the curve from to can be written as the sum of the area under the curve from to and from to , that is
Let us consider an example of this. Let
It is clear that the area under the curve from to , is .
Note that need not lie between and for this relation to hold true. Suppose that .
Analytically, this relation can be proved easily using the Newton Leibnitz’s formula.
(4) Let on the interval . Then,
This is because the curve of lies above the curve of , or equivalently, the curve of lies above the -axis for
Similarly, if on the interval , then
For the interval , suppose . That is, is a lower-bound for while is an upper bound. Then
This is obvious once we consider the figure below: