If ,
is a function on
, and
,
then the Riemann sum associated to these data is
By definition, the integral,
is the limit
(if it exists) of these Riemann sums as maximum of
tends
to zero.
(Applet illustrating Riemann sums)
For the sake of convenience, we often assume that the interval
has been decompose into
pieces of equal length, for some
positive integer
. The length of each piece is then
.
So,
and
and for such a
uniform decomposition, the Riemann sum is
Let ,
, and
be given. Then the lefthand Riemann sum approximating
with
subdivisions is given by setting
where
.
The righthand Riemann sum is given by setting
.
Compute the left and the right
approximations to
with
subdivisions.
In this case
. So,
The midpoint approximation (to the integral
with
subdivisions) is given by taking
to be the midpoint
of the interval
where
and
.
Thus,
and
Approximate
using the midpoint rule and
subdivisions.
Rather than approximating the area bounded by a function by rectangles, one
may use other shapes. For example, one may use trapezoids. The area of the trapezoid with
corners at ,
,
, and
is
.
The trapezoidal approximation to
with
subdivisions is
where
and
.
Compute the trapezoidal approximation to
with
subdivisions.
Simpson's rule for approximating integrals is based on approximating
by the area bounded by the parabola passing through
,
, and
.
Rather than deriving Simpson's rule from its geometric description, we write the Simpson's rule approximation in terms of the midpoint and trapezoidal approximations.
Directly,
Compute the Simpson's rule approximation to
with
subdivisions.
The exact value of
.
Approximations for .
Left | ![]() |
Midpoint rule | ![]() |
Simpson's rule |
![]() |
Exact value |
![]() |
Trapezoidal rule | ![]() |
Right | ![]() |
The error of an approximation is the absolute value of the
difference between
and the exact value of the itegral.
How many subdivisions do we need to guarantee that a midpoint approximation to
approximates
to two decimal points.
We need to have an error of less than . So, we want
We compute
and
. This function is decreasing
between
and
. So,
. So, we want
. Thus,
suffices.
We compute