188 CHAPTER 7. THE INTEGRAL
24. A differentiable function f defined on (0,∞) satisfies the following conditions.
f (xy) = f (x)+ f (y) , f ′ (1) = 1.
Find f and sketch its graph.
25. There is a general procedure for constructing methods of approximate integration.Consider [0,1] and divide this interval into n equal pieces determined by {x0, · · · ,xn}where xi − xi−1 = 1/n for each i. The approximate integration scheme for a functionf , will be of the form (
1n
) n
∑i=0
ci fi ≈∫ 1
0f (x) dx
where fi = f (xi) and the constants, ci are chosen in such a way that the above sumgives the exact answer for
∫ 10 f (x) dx where f (x) = 1,x,x2, · · · ,xn. When this has
been done, change variables to write∫ b
af (y) dy = (b−a)
∫ 1
0f (a+(b−a)x) dx
≈ b−an
n
∑i=1
ci f(
a+(b−a)(
in
))=
b−an
n
∑i=1
ci fi
where fi = f(a+(b−a)
( in
)). Show that when n = 1, you get an approximation
with trapezoids and when n = 2 you get an approximation with second degree poly-nomials. This is called Simpson’s rule. Show also that if this integration scheme isapplied to any polynomial of degree 3 the result will be exact. That is,
12
(13
f0 +43
f1 +13
f2
)=∫ 1
0f (x) dx
whenever f (x) is a polynomial of degree three. Show that if fi are the values of f ata, a+b
2 , and b with f1 = f( a+b
2
), it follows that the above formula gives
∫ ba f (x) dx
exactly whenever f is a polynomial of degree three.
26. Let f have four continuous derivatives on [xi−1,xi+1] where xi+1 = xi−1 + 2h andxi = xi−1 + h. Show using Problem 17, there exists a polynomial of degree three,p3 (x) , such that
f (x) = p3 (x)+14!
f (4) (ξ )(x− xi)4
Now use Problem 25 to conclude∣∣∣∣∫ xi+1
xi−1
f (x) dx−(
h fi−1
3+
h fi43
+h fi+1
3
)∣∣∣∣< M4!
2h5
5,
where M satisfies, M ≥ max{∣∣∣ f (4) (t)∣∣∣ : t ∈ [xi−1,xi]
}. You will approximate the
integral withm−1
∑i=0
(h f (x2ih)
3+
4h f (x2ih+h)
3+
h f(x(2i+2)h
)3
)