Example
12. Let
over
. Use
the Lower Riemann Sum to approximate the value of the integral.
Solution 12.
Remark. This will
be slow because many minimums need to be approximated.
You can skip the calculation and look at the recorded results. Change
4 to
20 to get all
of the table.
|
m sample points |
LowerRiemann[a,b,m] |
|
1 |
|
|
2 |
|
|
4 |
|
|
6 |
|
|
8 |
|
|
10 |
|
|
12 |
|
|
14 |
|
|
16 |
|
|
20 |
|
|
24 |
|
|
28 |
|
|
32 |
|
|
40 |
|
|
50 |
|
|
60 |
|
|
70 |
|
|
80 |
|
|
100 |
|
|
120 |
|
![[Graphics:../Images/RiemannSumMod_gr_251.gif]](../Images/RiemannSumMod_gr_251.gif)
![[Graphics:../Images/RiemannSumMod_gr_252.gif]](../Images/RiemannSumMod_gr_252.gif)
Animation 3. ( Lower Riemann Sum Lower Riemann Sum ). Internet hyperlink.
(c) John H. Mathews 2004