Calculus&Mathematica is pleased to acknowledge
the heavy influence of Emil Artin's book
Calculus with Analytic Geometry (notes by G. B. Seligman),
Committee on the Undergraduate Program,
Mathematical Asociation of America, 1957.
Although dated in spots, this short book remains a jewel.
Here is the idea of the integral:
Take a function
and plot it on an interval
.
It might look something like this:
If any picture is missing, then execute all the blank cells below.
![[Graphics:Images/index_gr_7.gif]](Images/index_gr_7.gif)
Look at the area between the graph of
and the
-axis:
![[Graphics:Images/index_gr_10.gif]](Images/index_gr_10.gif)
If a part of this area lies above the
-axis, then call it positive.
If a part lies below the
-axis, then call it negative.
This gives the signs as shown below:
![[Graphics:Images/index_gr_13.gif]](Images/index_gr_13.gif)
Now define a number that all folks call the integral of
from
to
.
You signify this number by writing
![]()
and you calculate its value by summing up the measurements of these areas taken with the corresponding signs.
Accordingly, if
are the points between
and
at which the curve crosses the
-axis as shown below:
![[Graphics:Images/index_gr_22.gif]](Images/index_gr_22.gif)
To get a quick review of this, grab all four plots and everything between them, animate, and run forward at speed 2 or 3.
Then you calculate
by taking the sum of four measurements:
![]()
where
and
are positive
and
and
are negative.
Given a specific function
![]()
and given specific numbers
and
with
,
when you calculate
,
do you get a number or do you get another function?
Make the indicated area measurements: