Math 128A, Spring 2007
Homework 5 Solution
5.2.8 (a) I(f) = Rb
af(x)dx ≈Rb
af((a+b)/2)dx = (b−a)f((a+b)/2) = M1(f).
(b) If we subdivide [a, b] into nsubintervals a=x0< x1<· · · < xn=b
with xi+1 =xi+hand apply M1to each subinterval, we get
I(f) = Zb
a
f(x)dx
=Zx1
x0
f(x)dx +· · · +Zxn
xn−1
f(x)dx
≈(x1−x0)f((x0+x1)/2) + · · · + (xn−xn−1)f((xn−1+xn)/2)
=hf(a+h/2) + · · · +hf (a+ (n−1/2)h)
(c) M1(1/(1 + x)) = 2/3≈.6667 and M2(1/(1 + x)) = 2/5 + 2/7≈.6857,
while I= log 2 ≈.6931.
5.2.10 (a) Following the derivation of Simpson’s rule,
I(f) = Zb
a
f(x)dx
≈Zb
a
P4(x)dx
=Zb
a
[f(x0)L0(x) + · · · +f(x4)L4(x)]dx
=C0f(x0) + · · · +C4f(x4)
where Ci=Rb
aLi(x)dx. These integrals are easily evaluated (but it is very
tedious!), and you get Boole’s rule.
(b) B4(1/(1+x)) = (2/45)∗(1/4)∗(7+32∗4/5+13∗2/3+32∗4/7+7∗1/2) ≈
.7006. This is just a little worse than M2(1/(1 + x)) from problem 8.
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