MISCELLANEOUS EXAMPLES ON CHAPTER VII 
295 
[We can choose k. so that Q n -<P n is of degree n-1: then 
and so 
- nP n )dx=0, f ß P n (Qn-<P n )dx=0, 
Now apply Ex. 39.] 
43. Approximate Values of definite integrals. Show that the error 
in taking - a) {<£ (a) + <£ (5)} as the vahie of the integral 1 (x) dx is less 
than ^M{b — a) 3 , where M is the maximum of | r//' (x) | in the interval {a, b); 
and the error in taking (b-a) <£{£(« + &)} 38 ^ ess than M{b — a) 3 . 
[Write f'{x)—(f){x) in Exs. 4, 5.] 
value is less than ^ sW M {b — a) 5 , where M is the maximum of <pW{x). 
[Use Ex. 6. This rule, which gives a very good approximation, is known as 
Simpson’s Rule. It amounts to taking one-third of the first approximation 
given above and two-thirds of the second.] 
Show that the approximation assigned by Simpson’s Rule is the area 
bounded by the lines x=a, x — b,y = 0 and a parabola with its axis parallel 
to OF and passing through the three points on the curve y = <\>(x) whose 
abscissae are a, \{a + b), b. 
It should be observed that if 4>{x) is any cubic polynomial, (f)W{x) = 0, 
and Simpson’s Rule is exact. That is to say, given three points whose 
abscissae are a, %{a + b), b, we can draw through them an infinity of curves 
of the type y=a+^x+yx 2 + 8x 3 ; and all such curves give the same area. For 
one curve 8 = 0, and this curve is a parabola, 
44. Apply Simpson’s Rule to the calculation of tt from the formula 
[The result is -7833... If we divide the integral into two, 
.2 
from 0 to % and | to 1, and apply Simpson’s Rule to the two integrals 
separately, we obtain -7853916 The correct value is -7853981....] 
{Math. Trip. 1903.) 
3 
46. Calculate the integrals 
to two places of decimals. [In the last integral the subject of integration is 
not defined for # = 0: but if we assign to it, when x=0, the value 1, it 
becomes continuous throughout the range of integration.]