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Unformatted text preview: 8.7 Numerical Integration Sometimes we cannot calculate an integral explicitly, but need to know its
numerical value. So we try to get good approximations to its actual value.
There are many ways to do this, and we shall outline some in this section. Let us assume that f is continuous on [0,61 and that we want to get an approximate values for
b
frmm
a We subdivide the interval [0, 6] into n equal intervals of length «bi—E. So we write
{(1, l = [$0,331] U [$1,$2l U [$2,273] U U [mu—1,33n] , where each interval [iii—1J5] has length x5 . "7 8
Let us approximate the area under the graph of J” from 13,1 to an, that is, let us approximate
a: i f(:1:)d.:z:. mi — 1
We present ﬁve different ways to do this. Below, the symbol 9: means approxi mately equal to.
(I) The Left Endpoint Rectangle Linen
f (331:4) ($i — {ti—1): f($iu1) 6—0:.
12 . £<KC~1> Ill 94 (II) The Right Endpoint Rectangle f fwdr
mi) (95,; — m) = f (we) b; (I; 112 ' Cm: (III) The Midpoint Rectangle 1'". f($)d$ 3i w1 f (xi—12+$1)($a—$s—1)=f (mi—12+mi) 5:10 llE 95 (IV) The Trapezoid 2:5 mm win1 51”. (mi—1) +f (2%)] (211' — $5_1) = % [f($i._1)+ f (3‘)] b II? (V) The Parabola
We ﬁnd a parabola, that' IS a quadratic polynomial y: 032 + bw—l— c, thatpws —0} through the points (maul, f (mg_1)), (M, 3' (M9), (3;, f (m;)). Then
we calculate the area. under that parabola, and use it as an approximation to 1:; ﬁx) 4m : :12“ “13MB mi—l Ell Now We add the approximations over each individual piece I r: :1 f (3;) dc: to get an approximation to the total integral 1: f (1:) (£91. When we add up all the
pieces, in each of the ﬁve methods above, we get: (I) Left Endpoint Rule b—a.
n Ln = [f(n=o) +f ($1) +f ($2) + + f ($124)]
(II) Right Endpoint Rule 13—
Rn:
’n’: (111) Midpoint Rule .. b —'a 220 + :21 11:1 + 2:2 2:2 + $3 1071—1 + urn
M n H 2 M 2 M 2 )+—+f(*~——2 )l (IV) 'Ikapezoidal Rule “ U ($1) + f (922) + 1‘ (mg) + + f (an b—e
T" 211. (V) Simpson’s Rule (from the Parabola) b fwd0) +f($n)
sn— 0. +2Ef(w1)+f(m2)++f(wn~1)]+ =6? +4“(20%)+r(ee)+r<ﬂea)+«+f(%ﬂ)l [r (ms) + 211991) + 2f (x2) + 2r (ms) + + 2r (m) + f (mun. Example
Approximate 21
m2:/ —d:z,
123 by applying the rules aboveto f (2:) = i on [(1, b] m [1,2] with n = 5.
Solution
Here Then our points are
6 7 8 9 
{$0,$1,$2,$3,$4,$5} 2 {113, 3:37 3:2}  Let’s substitute in the rules
(I) Left Endpoint Rule 97 L5 = b:Ef($o)+f($1)+f(w2)++f(w4)] g[we+f(;)+r(:>+f<%>] 1 5 5 5 5
_ E[1+E+?+§+§]_0'7456“' (11) Right. Endpoint Rule (2—0. 5 U (ml) +r (x2) + +r ($5)] .51— [1(3) HG) +1” (2) +1" (—2) +f(2)] 1 5 5 5 5 1
[ §+§+§+§]_0.6456m ll Rs 5 6
(III) Midpoint Rule Ms b:{rct’zwclm”(ﬁzzﬁgwm+434?»
he)+r<:—:>+rez)+f(%>+r<:—z>1 10 19 1” 104.3]:(15919... [ﬁ+13+ﬁ+ﬁ 19 H 1
3
1
3 (IV) Trapezoidal Rule T5 = 1’ I5“ [1‘ ($0) + 21‘ (x1) + 21132) + 21133) + 21' ($4) + 1‘ ($501  1 s 7 8 9 = 1.6[f(1)+2f(5—)+2f(g)+2f (5)+2f (3)+f(2)]
1 10 10 10 10 1 = E[1+_6,+_7.+§+3+§]=0.6956m (V) Simpson’s Rule 98_ 59
u 30 [ +2 ”(131)+f(m2)+f(x3)+f($4)]+
+4[11%)+f(ﬂ%ﬂ)+f(ﬂi§ﬂ)+f(£3%4)+f(n%a)] f (130) + 1’ ($5) ] 1 [ +2[f(‘"’) filwﬁeww i
: ——— — + .. _ _
3" we ﬂeaﬁe if aha2)]
1+l
= i +2[[g~+gig+g]] =06932
3“ +4[[i—2+i°+i—°+%9+i—°ll Note that on a calculator,
1112 = 0.69134... so Simpson’s Rule is best. Error Estimates
If we are going to use theﬁe rules, we need to have an wtimate of the error we incur when using them. In a similar way to the way we obtained the Lagrange
form of the remainder in Taylor polynomials, one can prove the following esti
mates: (I) Error in Trapezoidal Rule E: [bf(:r:)d1:—Tn b
= f m) an: — “’27: [f(wo) + 2f (55.) +21%) +2f ($3) + +2r(a:._1) + mo]
3 .
: —(61;T::) f”(C): for some c'between a, and I). Then if M is a number such that
[f"(:z:){ S M for all a: 6 [(1,1)] , we obtain (be af‘
13315 WM (II) Error in Simpson’s Rule 99 U]
in
IE n ff(w)dw—sn b _ b—a
j;f(x)dmw 6n _ (b — (1)5
2880n4 for some c between a and b. Then if M is a number such that f($o) + f(I'n)
+2 [f ($1) + f ($2) + + f (sun—1)] + +4 [feewrea—xaweea) ++f('~r;+%)] f“) (c), ‘f(4) ($) g Mfor allx E [0'35]: we obtain S (b — “)5
IE" g 2880n4 M" Example
Estimate the error in the Trapezoidal and Simpson’s rule for 2
1112 =/ ldaz.
1 m Solution
Here
I)" a = 2 — 1 = I,
and 1
f (x) = E:
so 2
f” (m) = ‘5 —> we» s 2,9: 6 [1,21 Then we have the bound M = 2 011 “ml, so Also 24 ,
f“) (m) = 5 => If“) (m) 5 24,21: 6 [1,2]. Then we have the bound M = 24 on I f(4) I, so 5
S < (b — a} = 1
”EM ‘ 2880214 24 120n4'
Clearly Simpson’s gives better accuracy, though it involves evaluating the func— tion at more points. . 100 ...
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 Spring '08
 Morley

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