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

17.2 Trapezium rule 499
The difference between the estimated value of the integral and the true value of the
integral istheerror. So
error = estimated value −truevalue
Weareabletofindthemaximumvalueof |error|.Firstlywecalculatethesecondderivative
of f, that is f ′′ . Suppose that |f ′′ | is never greater than some value,M, throughout
the interval [a,b], thatis
|f ′′ | M forallxvalues on [a,b]
WesaythatM isanupperboundfor |f ′′ |on[a,b].Thentheerrorduetothetrapezium
ruleissuch that
|error | (b−a) h 2 M
12
The expression (b−a) h 2 M is an upper bound for the error. Note that the error
12
depends upon h 2 : if the strip width, h, is halved the error reduces by a factor of 4; if
hisdivided by 10 the erroris divided by 100.
Example17.2 Find an upper bound for the error in the estimates calculated in Example 17.1. Hence
find upper and lower boundsforthe truevalue of ∫ 1.3
0.5 e(x2) dx.
Solution We have f = e (x2) . Then
f ′ = 2xe (x2 )
and f ′′ = 2e (x2) (1+2x 2 )
We note that f ′′ is increasing on [0.5,1.3] and so its maximum value is obtained at
x = 1.3. Thus the maximum value of f ′′ on [0.5,1.3] is 2e (1.3)2 [1 + 2(1.3) 2 ]. Noting
that2e (1.3)2 [1 +2(1.3) 2 ] = 47.47weseethat48isanupperboundfor f ′′ on[0.5,1.3],
thatisM = 48.
We notethata = 0.5,b= 1.3.Thus
|error | (0.8)h2 (48)
= 3.2h 2
12
(a) InExample 17.1(a),h = 0.2 and so
| error | 3.2(0.2) 2 = 0.128
Thus an upper bound for the error is0.128. We have
−0.128 error 0.128
The estimated value of the integral is2.117 and so
that is
2.117 −0.128 truevalue of integral 2.117 +0.128
1.989
∫ 1.3
0.5
e (x2) dx 2.245
An upper bound for ∫ 1.3
0.5 e(x2) dx is2.245 and a lower bound is 1.989.