Suppose is a non-negative, continuous function defined on some
interval
. Then by the area under the curve
between
and
we mean the area of the region bounded above by the
graph of
, below by the
axis, on the left by the vertical
line
, and on the right by the vertical line
.
All of the numerical methods in this lab depend on subdividing the
interval into subintervals of uniform length. For example,
dividing the interval
into four uniform pieces produces the
subintervals
,
,
, and
.
The Maple student package has commands for visualizing these three rectangular area approximations. To use them, you first must load the package via the with command. Then try the three commands given below. Make sure you understand the differences between the three different rectangular approximations. Take a moment to see that the different rules choose rectangles which in each case will either underestimate or overestimate the area.
> with(student):
> rightbox(x^2,x=0..4);
> leftbox(x^2,x=0..4);
> middlebox(x^2,x=0..4);
There are also Maple commands leftsum, rightsum, and middlesum to sum the areas of the rectangles, see the examples below. Note the use of evalf to obtain numerical answers.
> rightsum(x^2,x=0..4);
> evalf(rightsum(x^2,x=0..4));
> middlesum(x^2,x=0..4);
> evalf(middlesum(x^2,x=0..4));
It should be clear from the graphs that adding up the areas of the
rectangles only approximates the area under the curve. However, by
increasing the number of subintervals the accuracy of the
approximation can be increased. All of the Maple commands described so
far in this lab permit a third argument to specify the number of
subintervals. The default is 4 subintervals. See the example below
for appoximations to the area under from
to
using
the rightsum
command with 4, 10, 20 and 100 subintervals. (As this region describes
a right triangle with height 2 and base 2, this area can be easily
calculated to be exactly 2.) Try it yourself with the leftsum
and middlesum commands.
> evalf(rightsum(x,x=0..2));
> evalf(rightsum(x,x=0..2,10));
> evalf(rightsum(x,x=0..2,20));
> evalf(rightsum(x,x=0..2,100));
Since, in this trivial example, we knew that the area is exactly 2, it appears that, as the number of subintervals increases, the rectangular approximation becomes more accurate. What would happen with a not so trivial region? The next section describes a way of assessing the accuracy of the midpoint rule.
where is the absolute maximum of
on
,
assuming that
is continuous on that interval. In
practice,
is often approximated by a number
that is an upper bound
for
, that is
. For instance, if
on
,
might be taken as 4. Do you see why? For more
complicated functions, Maple can be used to get a value for
that is
close to or actually equal to
.
Note that the error bound formula gives a worst case estimate, the
accuracy achieved for a given number of subintervals
may be much
better than the guarantee
given by the formula.