Algebra Integrals

CHAPTER 1

Integrals


1.1. Areas and Distances. The Definite Integral

1.1.1. The Area Problem. The Definite Integral. Here we
try to find the area of the region S under the curve y = f (x) from a to
b, where f is some continuous function.

Y




y=f(x)




S


O a X
b


In order to estimate that area we begin by dividing the interval
[a, b] into n subintervals [x0 , x1 ], [x1 , x2 ], [x2 , x3 ], . . . , [xn−1 , xn ], each
of length ∆x = (b − a)/n (so xi = a + i∆x).

Y




O a X
b

6

, 1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 7

The area Si of the strip between xi−1 and xi can be approximated
as the area of the rectangle of width ∆x and height f (x∗i ), where x∗i
is a sample point in the interval [xi , xi+1 ]. So the total area under the
curve is approximately the sum
Xn
f (x∗i ) ∆x = f (x∗1 ) ∆x + f (x∗2 ) ∆x + · · · + f (x∗n ) ∆x .
i=1

This expression is called a Riemann Sum.
The estimation is better the thiner the strips are, and we can iden-
tify the exact area under the graph of f with the limit:
Xn
A = lim f (x∗i ) ∆x
n→∞
i=1

As long as f is continuous the value of the limit is independent of the
sample points x∗i used.
Rb
That limit is represented a f (x) dx, and is called definite integral
of f from a to b:
Z b Xn
f (x) dx = lim f (x∗i ) ∆x
a n→∞
i=1

The symbols at the left historically were intended to mean
R an infinite
sum, represented by a long “S” (the integral symbol ), of infinitely
small amounts f (x) dx. The symbol dx was interpreted as the length of
an “infinitesimal” interval, sort of what ∆x becomes for infinite n. This
interpretation was later abandoned due to the difficulty of reasoning
with infinitesimals, but we keep the notation.
Remark : Note that in intervals where f (x) is negative the graph of
y = f (x) lies below the x-axis and the definite integral takes a negative
value. In general a definite integral gives the net area between the
graph of y = f (x) and the x-axis, i.e., the sum of the areas of the
regions where y = f (x) is above the x-axis minus the sum of the areas
of the regions where y = f (x) is below the x-axis.

1.1.2. Evaluating Integrals. We will soon study simple and ef-
ficient methods to evaluate integrals, but here we will look at how to
evaluate integrals directly from the definition.
R1
Example: Find the value of the definite integral 0 x2 dx from its
definition in terms of Riemann sums.

, 1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 8

Answer : We divide the interval [0, 1] into n equal parts, so xi = i/n
and ∆x = 1/n. Next we must choose some point x∗i in each subinterval
[xi−1 , xi ]. Here we will use the right endpoint of the interval x∗i = i/n.
Hence the Riemann sum associated to this partition is:
X n µ ¶2
1 X 2
n
i 1 2n3 + 3n2 + n 2 + 3/n + 2/n2
1/n = 3 i = 3 = .
i=1
n n i=1 n 6 6

So:
Z 1
2 + 3/n + 2/n2 1
x2 dx = lim = .
0 n→∞ 6 3

In order to check that the result does not depend on the sample
points used, let’s redo the computation using now the left endpoint of
each subinterval:
Xn µ ¶2
1 X
n
i−1 1 2n3 − 3n2 + n 2 − 3/n + 2/n2
1/n = 3 (i−1)2 = 3 = .
i=1
n n i=1 n 6 6

So:
Z 1
2 − 3/n + 2/n2 1
x2 dx = lim = .
0 n→∞ 6 3


1.1.3. The Midpoint Rule. The Midpoint Rule consists of com-
puting Riemann sums using xi = (xi−1 + xi )/2 = midpoint of each
interval as sample point. This yields the following approximation for
the value of a definite integral:
Z b Xn
f (x) dx ≈ f (xi ) ∆x = ∆x [f (x1 ) + f (x2 ) + · · · + f (xn )] .
a i=1


R1
Example: Use the Midpoint Rule with n = 5 to approximate 0
x2 x.

Answer : The subintervals are [0, 0.2], [0.2, 0.4], [0.4, 0.6], [0.6, 0.8],
[0.8, 1], the midpoints are 0.1, 0.3, 0.5, 0.7, 0.9, and ∆x = 1/5, so
Z 1
1 £ 2 ¤
x2 dx ≈ 0.1 + 0.32 + 0.52 + 0.72 + 0.92 = 1.65/5 = 0.33 ,
0 5
which agrees up to the second decimal place with the actual value 1/3.

, 1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 9

1.1.4. The Distance Problem. Here we show how the concept of
definite integral can be applied to more general problems. In particular
we study the problem of finding the distance traveled by an object with
variable velocity during a certain period of time.
If the velocity v were constant we could just multiply it by the time
t: distance = v × t. Otherwise we can approximate the total distance
traveled by dividing the total time interval into small intervals so that
in each of them the velocity varies very little and can can be considered
approximately constant. So, assume that the body starts moving at
time tstart and finishes at time tend , and the velocity is variable, i.e., is
a function of time v = f (t). We divide the time interval into n small
intervals [ti−1 , ti ] of length ∆t = (tend − tstart )/n, choose some instant t∗i
between ti−1 and ti , and take v = f (t∗i ) as the approximate velocity of
the body between ti−1 and ti . Then the distance traveled during that
time interval is approximately f (t∗i ) ∆t, and the total distance can be
approximated as the sum
Xn
f (t∗i ) ∆t
i=1

The result will be more accurate the larger the number of subintervals
is, and the exact distance traveled will be limit of the above expression
as n goes to infinity:
Xn
lim f (t∗i ) ∆t
n→∞
i=1
That limit turns out to be the following definite integral:
Z tend
f (t) dt
tstart


1.1.5. Properties of the Definite Integral.
Z b
(1) Integral of a constant: c dx = c (b − a).
a
(2) Linearity:
Z b Z b Z b
(a) [f (x) + g(x)] dx = f (x) dx + g(x) dx.
a a a
Z b Z b
(b) cf (x) dx = c f (x) dx.
a a
(3) Interval Additivity