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

504 Chapter 17 Numerical integration
Example17.6 Find an upper bound for the error in the estimates calculated in Example 17.3. Hence
find upper and lower bounds for ∫ 1.3
0.5 e(x2) dx.
Solution Here wehave f = e (x2) . Calculating the fourthderivative gives
f (4) = 4e (x2) (4x 4 +12x 2 +3)
Weseekanupperboundfor |f (4) |on[0.5,1.3].Wenotethat f (4) increasesasxincreases
and soits maximum value inthe interval occurs whenx = 1.3:
f (4) (1.3) = 752.319 < 753
Hence 753 is anupper bound for |f (4) |.
Inthis examplea = 0.5 andb=1.3 and so
| error | (0.8)h4 (753)
180
(a) Hereh = 0.2 and so
= 3.347h 4
|error | 3.347(0.2) 4 = 0.0054
An upper bound forthe erroris0.0054. Now
−0.0054 error 0.0054
The estimated value of the integral is2.0762 and so
2.0762 −0.0054
∫ 1.3
0.5
e (x2) dx 2.0762 +0.0054
thatis
2.0708
∫ 1.3
0.5
e (x2) dx 2.0816
(b) Hereh = 0.1 and so
|error | 3.347(0.1) 4 = 3.347 ×10 −4
An upper bound forthe erroris3.347 ×10 −4 . Now
−3.347 ×10 −4 error 3.347 ×10 −4
Noting thatthe estimated value of the integral is2.0749 we have
2.0749 −3.347 ×10 −4
∫ 1.3
0.5
e (x2) dx 2.0749 +3.347 ×10 −4
thatis
2.0746
∫ 1.3
0.5
e (x2) dx 2.0752