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

17.2 Trapezium rule 497
y
{ {
y 0 y n
y i y i +1
x i
a
x i +1 b
h
{
{
x
Figure17.1
Each stripis approximated by atrapezium.
If the number of strips is increased, that ishis decreased, then the accuracy of the approximation
is increased.
Table17.1
Example17.1 Usethe trapezium ruletoestimate
(a) a strip widthof0.2
∫ 1.3
0.5
e (x2) dx using
(b) a strip widthof0.1.
Solution (a) Table 17.1 lists values ofxand correspondingvalues ofe (x2) .
We notethath = 0.2.Usingthe trapeziumrulewefind
x
0.5
y=e (x2 )
1.2840 =y 0
sum ofareasoftrapezia = 0.2 {1.2840 +2(1.6323) +2(2.2479)
2
+2(3.3535) +5.4195}
0.7 1.6323 =y 1
= 2.117
0.9 2.2479 =y 2
Table17.2
1.1 3.3535 =y 3 Hence
1.3 5.4195 =y 4 ∫ 1.3
x y=e (x2 )
0.5
e (x2) dx ≈ 2.117
x 0 = 0.5 1.2840 =y 0
x 1 = 0.6 1.4333 =y 1
x 2 = 0.7 1.6323 =y 2
x 3 = 0.8 1.8965 =y 3
x 4 = 0.9 2.2479 =y 4
x 5 = 1.0 2.7183 =y 5
x 6 = 1.1 3.3535 =y 6
x 7 = 1.2 4.2207 =y 7
x 8 = 1.3 5.4195 =y 8
(b) Table 17.2 listsxvalues and correspondingvalues ofe (x2) .
Using Table17.2, wefind
Hence
sum ofareasoftrapezia = 0.1 {1.2840 +2(1.4333) +2(1.6323) + ···
2
+2(4.2207) +5.4195}
∫ 1.3
0.5
e (x2) dx ≈ 2.085.
= 2.085
Dividingtheinterval[0.5,1.3]intostripsofwidth0.1resultsinamoreaccurateestimate
thanusingstripsofwidth0.2.