082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

17.3 Simpson’s rule 503
Table17.6
Time(s) Voltage(V) Current(A)
0 50.5 10.1
10 101.0 20.2
20 67.5 13.5
30 80.5 16.1
40 92.0 18.4
50 96.0 19.2
60 78.5 15.7
70 82.0 16.4
80 90.5 18.1
90 107.0 21.4
100 86.0 17.2
Table17.7
Time(s) Voltage(V) Current(A) Power(W)
0 50.5 10.1 510.05
10 101.0 20.2 2040.20
20 67.5 13.5 911.25
30 80.5 16.1 1296.05
40 92.0 18.4 1692.80
50 96.0 19.2 1843.20
60 78.5 15.7 1232.45
70 82.0 16.4 1344.80
80 90.5 18.1 1638.05
90 107.0 21.4 2289.80
100 86.0 17.2 1479.20
wherePis the power. AlsoP =IV, and so
E =
∫ t
0
IV dt
We first need toevaluatePas shown inTable 17.7.
Then using Simpson’s rulewithh = 10 wehave
E =
∫ 100
0
IVdt≈ 10 {510.05 +4(2040.20 +1296.05 +1843.20 +1344.80
3
+2289.80) +2(911.25 +1692.80 +1232.45 +1638.05)
+1479.20}
= 160648.5 J
= 160.649 kJ
The energy dissipated istherefore approximately 160.6 kJ.
17.3.1 ErrorduetoSimpson’srule
Simpson’sruleprovidesanestimatedvalueofadefiniteintegral.Thedifferencebetween
theestimatedvalueandthetrue(exact)valueistheerror.Justaswiththetrapeziumrule,
wecan calculateanupper bound forthiserror.
We need to calculate the fourth derivative of f, that is f (4) . Suppose |f (4) | is never
greaterthansome value,M, throughoutthe interval [a,b], thatis
|f (4) | M forallxvalues on [a,b]
ClearlyM isanupperboundfor |f (4) |on[a,b].TheerrorduetoSimpson’sruleissuch
that
|error | (b−a)h4 M
180
Note that the error isproportional toh 4 .