Thomas Calculus 13th [Solutions]

Section 8.7 Numerical Integration 615
26. 24
2
0.019 2(0.020) 2(0.021) 2(0.031) 0.035 4.2 L
27. (a)
(b)
b a 4
2
Es
x M ; n 4 x
0 ;
180
4 8
x x
8 3 24 ;
mf ( x i ) 10.47208705
S (10.47208705) 1.37079
24
(c) 0.00021 100 0.015%
1.37079
(4)
2
f 1 M 1 E 0 4
s
(1) 0.00021
180 8
x i f ( xi
) m mf ( x1i
)
x0
0 1 1 1
x 1 /8 0.974495358 4 3.897981432
x 2 /4 0.900316316 2 1.800632632
x3
3 /8 0.784213303 4 3.136853212
x 4 /2 0.636619772 1 0.636619772
28. (a) x b a 1 0 0.1 erf (1) 2 0.1 y
10 3 3 0 4y1 2y2 4y3 4y n
9 y10
(b)
2
30
E s
0 0.01 0.04 0.09 0.81 1
e 4e 2e 4e 4e e 0.843
1 0 4 6
(0.1) (12) 6.7 10
180
29. T x y0 2y 2 1 2y2 2y3 2yn
1 y n where x b a and f is continuous on [ a, b ]. So
n
b a
n
y y y y y y y y b a f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x )
2 n 2 2 2
T 0 1 1 2 2 n 1 n 1 n 0 1 1 2
n 1 n
. Since f is
continuous on each interval xk
1 , x k , f ( xk
1) f ( xk
)
and is always between f ( x
2
k 1) and f ( x k ), there is
a point c k in k 1 ,
f ( xk
1) f ( xk
)
x x k with f ( c k ) ; this is a consequence of the Intermediate Value Theorem.
2
n
n
Thus our sum is b a f ( c k ) which has the form x
n
k f ( c k ) with x b a
k for all k. This a Riemann
n
k 1
k 1
Sum for f on [ a, b].
30. S x y0 4y 3 1 2y2 4y3 2yn 2 4yn 1 y n where n is even, x b a and f is continuous on
n
y0 4 y1 y2 y2 4 y3 y4 y4 4 y5 y6 yn 2 4 yn 1 yn
[ a, b ]. So S b a
n 3 3 3 3
b a
n
2
f ( x0 ) 4 f ( x1 ) f ( x2 ) f ( x2 ) 4 f ( x3 ) f ( x4 ) f ( x4 ) 4 f ( x5) f ( x6 ) f ( xn 2 ) 4 f ( xn 1) f ( xn
)
6 6 6 6
f ( x2k ) 4 f ( x2k 1) f ( x2k
2)
is the average of the six values of the continuous function on the interval
6
x2k
, x 2k
2 , so it is between the minimum and maximum of f on this interval. By the Extreme Value
Theorem for continuous functions, f takes on its maximum and minimum in this interval, so there are x a and
f ( x2k ) 4 f ( x2k 1) f ( x2k
2)
x b with x2k xa , xb x 2k
2 and f ( xa
) f ( x ).
6
b
f ( x2k ) 4 f ( x2k 1) f ( x2k
2 )
By the Intermediate Value Theorem, there is c k in x2k
, x 2k
2 with f ( c k ) .
6
n/2
So our sum has the form xk
f ( c k ) with x b a
k ( /2) , a Riemann sum for f on [ a, b].
n
k 1
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