Integral Calculus By Jorge Samayoa

§1. (Short)History The history of calculus stares several hundreds of years ago, when the mathematician

Archimedes (225 B.C.) made a very important contribution. It said that the area of a

segment of a parabola is 43

the area of a triangle with the same base and vertex and 23

the area of the circumscribed parallelogram. In his work, Archimedes constructed an

infinite series of triangles of known area to approximate the area of such segment.

Archimedes, therefore, introduces the concept of putting together different pieces of

known area to get a different area. This seems to be a very basic thought, but it was very

influential to the basis of calculus.

Several years after Archimedes work, and several contributions of different people, there

were two people that took, pretty much, the whole credit of what is know as calculus.

These people were, Sir Isaac Newton and Gottfried Leibniz. Even thought the first person

who gives a precise idea of calculus was Newton, the one that publish first his work

know as calculus was Leibniz. At that time, there was a controversy of who invented

calculus first and which country deserved the credit. A careful examination of the papers

of Leibniz and Newton shows that they arrived at their results independently, with

Leibniz starting first with integration and Newton with differentiation. Today, both

Newton and Leibniz are given credit for developing calculus independently. It is Leibniz,

however, who gave the new discipline its name. Newton called his calculus "the science

of fluxions".

Nowadays, independently of whose ideas we should follow, calculus is divided in two

huge problems: The first problem of calculus, which is the estimation of the slope of a

curve in a given point, and the second problem of calculus which is the estimation of the

area under a curve in a given interval. In this document, however, we will focus in the

second problem of calculus since we already went over the first problems in the last

tutorial (The Geometric Approach of the Derivative).

§2. The Second Problem of Calculus (The Area Problem)

(http://calclab.math.tamu.edu/~jsamayoa/final_section_2 ) The second problem of calculus is finding the area of the region R that lies under the

curve ( )y f x= from a to b (Figure 1).

Figure 1: Area under the curve

In order to find the area under a curve, as described in the section below, we will

approximate the area by using another figure of known area; the rectangle (Figure 2).

Figure 2: Area of the Rectangle

One of the persons who introduced this was Jean-Gaston Darboux. However, this

approach is equivalent to the approach given by Friedrich Bernhard Riemann.

There are two different ways of doing this approximation. One is the Darboux upper

sums, figure 3, and the Darboux lower sums, figure 4. As you might notice, the

difference relies on the selection of the lengths of the approximating rectangles.

1

0

[ , ]

1

Let :[ , ] be a bounded function, and let ( , , ) be a partition of [ , ]. Let

sup ( ).

The upper Darboux sum of with respect of is

U[ , ] .

i i

n

ix x x

n

i ii

f a bP x x a b

M f x

f P

f P M x

−∈

=

→=

=

= Δ∑

…

1where ( ).i i ix x x −Δ = −

Figure 3: Upper Darboux sum.

1

0

[ , ]

1

1

Let :[ , ] be a bounded function, and let ( , , ) be a partition of [ , ]. Let

inf ( ).

The lower Darboux sum of with respect of is

[ , ] .

where ( ).

i i

n

i x x x

n

i ii

i i i

f a bP x x a b

m f x

f P

L f P m x

x x x

−∈

=

−

→=

=

= Δ

Δ = −

∑

…

Figure 4: Lower Darboux sum.

Theorem 1:

Let be the area of the region R that lies under the curve ( ) from to , and let be a Partition of [ , ]. Then, [ , ] [ , ]

A y f x a b Pa b

L f P A U f P

=

≤ ≤

Proof: (We let the proof of this result to the reader).

In order to improve our approximation of the area A. Let us make a refinement of P0 in

the Darboux upper sums, say P1. Notice that the difference between the approximation A

using P0 and using P1 is the length of the interval 1[ , ]i ix x − .

If we refine Pi by Pj (i

Definition 2:

Let Ai be the area of the ith approximating rectangle. Then, the area A of the region R

that lies under the graph of [ , ]f C a b∈ is

1 1 2 2lim lim[ ( ) ( ) ( ) ]i n nn nA A f x x f x x f x x→∞ →∞= = Δ + Δ + + Δ

for [ , ]jx a b∈ .

Example 1:

Estimate the area under the curve 2y x= for [0,1]x∈ by

1. 4 approximating rectangles by

1.1. Darboux upper sums.

1.2. Darboux lower sums.

2. n approximating rectangles, then, let n →∞ .

SOLUTION:

1.1

( )

14

1 1 3 14 2 4

1 1 1 9 1 0.46884 16 4 16

b axn

A f x f x f x f x

−Δ = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞≈ Δ + Δ + Δ + Δ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞≈ + + + =⎜ ⎟⎝ ⎠

1.2

( )

14

1 1 304 2 4

1 1 1 9 0 0.21884 16 4 16

b axn

A f x f x f x f x

−Δ = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞≈ Δ + Δ + Δ + Δ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞≈ + + + =⎜ ⎟⎝ ⎠

2

( )

( )

2 2 2

2 2 22 2

1

1 2 1

1 1 2

1 1 ( 1)(2 1) 1 26

( 1)(2 1) 1lim .6 3n

b axn n

A f x f x f xn n

nn n n n

n n nnn n

n nAn→∞

−Δ = =

⎛ ⎞ ⎛ ⎞≈ Δ + Δ + + Δ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞≈ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

+ +⎛ ⎞≈ + + + = ⎜⎝ ⎠

+ +⎛ ⎞∴ = =⎜ ⎟⎝ ⎠

§3. Definite Integral.

(http://calclab.math.tamu.edu/~jsamayoa/final_section_3 )

Definition 3:

Suppose [ , ]f C a b∈ .Let us divide the interval [ , ]a b in n subintervals of width

b axn−

Δ = . Let 0 1 2, , , , nx a x x x b= =… , be the endpoints and * * * *0 1 2, , , , nx x x x… the sample

points in such intervals, so * 1 lies in the th subinterval [ , ].i i ix i x x− Then the definite

integral of f from a to b is

( )*1

( ) limb n

in ia

f x dx f x x→∞

=

= Δ∑∫

Since we supposed that f is continuous in [a , b], definition 3 can be states as follows.

For every number 0ε > , there is a positive integer number N such that,

( )*1

( )b n

iia

f x dx f x x ε=

− Δ N and for every * 1in [ , ].i i ix x x−

Example 2:

Evaluate 1

0

.xdx∫

SOLUTION:

( )

( )

( )

1*

10

1

0

2 2

By definition ( ) lim .

1Now, the width is .

thus,

1 2 1

1 1 2

1 1 ( 1) 1 22

n

in i

f x dx f x x

b axn n

xdx f x f x f xn n

nn n n n

n nnn n

→∞=

= Δ

−Δ = =

⎛ ⎞ ⎛ ⎞≈ Δ + Δ + + Δ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞≈ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠+⎛ ⎞≈ + + + = ⎜ ⎟

⎝ ⎠

∑∫

∫

1

0

( 1) 1 lim .2 2n

nxdxn→∞+⎛ ⎞∴ = =⎜ ⎟

⎝ ⎠∫

§3.1. Properties of the definite integral.

Due the nature of this document, we will state some of the most important properties of

the definite integral, and the proofs of such properties are lived as exercise for the

student.

Properties of the Integral.

1. ( ), where is any constant.b

a

cdx c b a c= −∫

2. ( ) ( ) ( ) ( )b b b

a a a

f x g x dx f x dx g x dx± = ±∫ ∫ ∫

3. ( ) ( ) , where is any constant.b b

a a

cf x dx c f x dx c=∫ ∫

4. ( ) ( ) ( )b c c

a b a

f x dx f x dx f x dx+ =∫ ∫ ∫ with ( , )c a b∈ .

5. If ( ) for , thenm f x M a x b≤ ≤ ≤ ≤

( ) ( ) ( ).b

a

m b a f x dx M b a− ≤ ≤ −∫

Example 3:

Use properties of the integral to evaluate 1

0

(2 4 ) .x dx+∫

1 1 1 1

0 0 0 0

1

0

(2 4 ) 2 4 2(1 0) 4

by example 2,

1(2 4 ) 2 4 42

x dx dx xdx xdx

x dx

+ = + = − +

⎛ ⎞+ = + =⎜ ⎟⎝ ⎠

∫ ∫ ∫ ∫

∫

Definition 4 (Fundamental Theorem of Calculus) Suppose [ , ]f C a b∈ . Then,

1. If ( ) ( ) , then '( ) ( ).x

a

g x f t dt g x f x= =∫

2. ( ) ( ) ( ), where is any antiderivative of , that is, ' .b

a

f x dx F b F a F f F f= − =∫

Proof: Since this is a support document for out tutorial, the proof of this will be in other

document with different goal.

§4. Indefinite Integrals.

(http://calclab.math.tamu.edu/~jsamayoa/final_section_4 ) The goal of this section is to familiarize the student with the evaluation of indefinite

integrals. Therefore, we will just present a table with several indefinite integrals so that

the student might have a comprehension of the computational procedure of solving

indefinite integrals.

Because of the relation given by the Fundamental Theorem of Calculus between

antiderivatives and integrals, the notation ( )f x dx∫ is traditionally used for an antiderivatives of f and is called indefinite integral. Thus,

( ) ( ) means '( ) ( )f x dx F x F x f x= =∫

For example, we can write 2 2

, because 2 2x d xxdx C C x

dx⎛ ⎞

= + + =⎜ ⎟⎝ ⎠

∫ .

Table of Integrals.

Example 4:

a) ( )5

43 cos 3 sin5xx x dx x C+ = + +∫

b) 22tan sec 2lnx dx x x Cx

⎛ ⎞+ = + +⎜ ⎟⎝ ⎠∫

c) 12

4 4sin1

e d e Cθ θθ θθ

−⎛ ⎞+ = + +⎜ ⎟−⎝ ⎠

∫

d) Use the Fundamental Theorem of Calculus, to

evaluate2

42

0

22 3 .1

x x dxx

⎛ ⎞+ −⎜ ⎟+⎝ ⎠∫

22 5 24 1

20 0

5 21

1

22 3 2 3 2 tan1 5 2

2 2 2 3 2 tan 2 05 2

64 6 2 tan 25

94 25

x xx x dx xx

−

−

−

⎛ ⎞+ − = + −⎜ ⎟+⎝ ⎠

⎛ ⎞= + − −⎜ ⎟⎝ ⎠

= + −

= −

∫

1tan 2−

§5 Area between two curves.

(http://calclab.math.tamu.edu/~jsamayoa/final_section_5 ) In section 2 we defined and calculated areas of regions that lie under curves. In this

section we will use integrals to find areas of region between two curves.

In other words, our gold in this section is to calculate the area of the region R defined by

the graphs of two functions , [ , ]f g C a b∈ , such that y=f(x) and y=g(x) for

a x b≤ ≤ (figure 6).

Figure 6: Area between Curves.

In order to find a precise formula for calculating area between curves, let us follow the

same procedure we used in section 2 for finding areas of regions under curves.

First, we calculate the differential of area Ai, then we approximate the area between the

curves by the sum of the Ai’s. The following figure (figure 7) shows how the differential

of area is calculated.

Figure 7: Analyzing Area between curves.

Now, we sum the Ai’s to get,

( )* *1 1

( ) ( )n n

i i ii i

A A f x g x x= =

≈ = − Δ∑ ∑

Let ( 0)n x→∞ Δ → to get

( )* *1

lim ( ) ( )n

i in iA f x g x x

→∞=

= − Δ∑

Definition 5:

The area A of the region bounded by the functions , [ , ]f g C a b∈ , such that y=f(x) and

y=g(x) with ( ) ( )f x g x≥ for a x b≤ ≤ , is

( )( ) ( )b

a

A f x g x dx= −∫

Example 5:

Find the area of the region R enclosed by 2 1and 2 with 0, .2

xy e y x x ⎡ ⎤= = − + ∈⎢ ⎥⎣ ⎦

SOLUTION: (Check the tutorial solution for this problem).

( )1

22

01

3 2

03

12

12

( 2)

23

112 2 1

3 21 2 0.309624

x

xx

x

A x e dx

x x e

e

e

=

=

= − + −

= − + −

⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠= − + − −⎜ ⎟

⎝ ⎠

= − − + ≈

∫

Example 6:

Find the area of the region R enclosed by 2 2( ) and ( ) 2 .f x x g x x x= = −

SOLUTION: (Check the tutorial solution for this problem)

First we find the intersection point of these curves. 2 2 2( ) ( ) 2 2 2 0 2 ( 1) 0

0 and 1.f x g x x x x x x x x

x x= ⇒ = − ⇒ − = ⇒ − =

∴ = =

Secondly, we integrate de difference of the functions from 0 to 1, as in definition 5.

Notice that in this example ( ) ( ) for [0,1].g x f x x≥ ∈ Thus

( )

( )

( )

1

01

2

01

2

012 3

0

( ) ( )

2 2

2

22 3

1 1 12 .2 3 3

g x f x dx

x x dx

x x dx

x x

−

= −

= −

⎡ ⎤= −⎢ ⎥

⎣ ⎦

⎛ ⎞= − =⎜ ⎟⎝ ⎠

∫

∫

∫

References:

• Calculus: Early Transcendentals, James Steward, 6th edition. Brooks Cole.

• Calculus: Concepts and Contexts , James Steward, Brooks Cole.

• Calculus of a Single Variable, Larson – Hostetler-Edwards, 8th edition,

Houghton Mifflin Company.

• Calculo Diferencial e Integral, Stefan Banach, 2nd. Edition, Uthea.

• Wikipedia: Bernhard Reimann.

• Wikipedia: Jean Gaston Darboux. < http://en.wikipedia.org/wiki/Darboux>