1

Let us consider the integral of from . To evaluate the area under , we can separately evaluate the area under and the area under and add the two areas (algebraically). Thus:

Now consider the integral of from to . To evaluate the area under , we can first evaluate the area under and then multiply it by , that is:

2

Consider an odd function , i.e., . This means that the graph of is symmetric about the origin.

From the figure, it should be obvious that because the area on the left side and that on the right algebraically add to .

Similarly, if was even, i.e.,

because the graph is symmetrical about the -axis.

If you recall the discussion in the unit on functions, a function can also be even or odd about any arbitrary point . Let us suppose that is odd about , i.e

Suppose for example, that we need to calculate . It is obvious that this will be , since we are considering equal variation on either side of , i.e. the area from to and the area from to will add algebraically to .

Similarly, if is even about , i.e.

From this discussion, you will get a general idea as to how to approach such issues regarding even/odd functions.

**(3)**Let us consider a function on

We want to somehow define the “average” value that takes on the interval . What would be an appropriate way to define such an average?

Let be the average value that we are seeking. Let it be such that it is obtained at some

We can measure by saying that the area under from to should equal the area under the average value from to . This seems to be the only logical way to define the average (and this is how it is actually defined!). Thus

This value is attained for at least one (under the constraint that is continuous, of course).