The Art and Craft of Problem Solving

336 CHAPTER 9 CALCULUS

for all x 2 o. This certainly implies that [(x)^2 � [(Of + [' (0)^2. Thus there is a

constant

C := J [(0)^2 + [' (0)^2 ,

for which I[(x) I � C for all x 2 o. We will be done if we can do the argument for

x < o. We leave that as an exercise.

Integration

The fundamental theorem of calculus gives us a method for computing definite inte­

grals. We are not concerned here with the process of antidifferentiation-we assume

that you are well versed in the various techniques-but rather with a better understand­

ing of the different ways to view definite integrals. There is a lot of interplay among

summation, integration, and inequalities; many problems exploit this.

n n

Example 9.3.8 Compute lim � -

k2 2

n.

--->oogl +n

Solution: The problem is impenetrable until we realize that we are faced not with

a sum, but the limit of a sum, and that is exactly what definite integrals are. So let us

try to work backwards and construct a definite integral whose value is the given limit.

Recall that we can approximate the definite integral l

b

[(x)dx by the sum

1

Sn := -(/(a) + [(a +L1) + [(a+ 2L1) + ... + [(b -L1)),

n

where L1 =

b -a

. Indeed, if [ is integrable,
n

a

lim Sn = r

b

[(x)dx.

n ---+oo Ja

b

n

Now it is just a matter of getting �

k^2

n

2

to look like Sn for appropriately

gl +n

chosen [(x),a, and b. The crux move is to extract something that looks like L1 =

(b -a) / n. Observe that