Introduction to Integration

Area and Definite Integral

Chapter 4. INTEGRALS

4.1 Area problem

• We know how to compute areas of rectilinear objects, such as rectangles, triangles, polygons

• How do we define and compute areas of more complicated regions (e.g. area enclosed by a circle)?

• Idea: approximate such regions by rectilinear regions (for example, by polygons)

Area under the curve y=f(x) between a and b

x

y

a

y = f(x)

bx

Assume f(x) ≥0 on [a,b] and consider regionR = { (x,y) | a ≤x ≤ b, 0 ≤ y ≤ f(x) }

(x,y)

f(x)

What is the area of R?

=xnx1 x2xi-1 xi

Approximation by rectangles

x

y

a

y = f(x)

bx0=

• Divide [a,b] into n intervals of equal length

• Use right endpoints to built rectangles (columns)

Area of i-th column is f(xi)•∆x

x

y

a

y = f(x)

bx0= =xnx1 x2xi-1 xi

f(xi)

∆x

f(xi)f(xi)∆x

Total area of all columns is

x

y

a

y = f(x)

bx0= =xnx1 x2xi-1 xi

f(xi)

∆x

f(xi)∆x

f(x2)∆xf(x1)∆x

f(xn)∆x

xxfxxfxxfxxf ni )(...)(...)()( 21

xxfRn

iin

1

)(

Definition. Area under the curve is

x

y

a

y = f(x)

b

xxfRAn

ii

nn

n

1

)(limlim

n=14

Theorem. If f is continuous on [a,b] then the following limit exists:

x

y

a

y = f(x)

b

xxfAn

ii

n

1

)(lim

How to find xi

x

a bx0= =xnx1 x2xi-1 xi

∆x ∆x ∆x

nabaxaxxax

nabaxaxax

axoax

/)(22

/)(1

2

1

0

nabiaxiaxi /)(

nabx /)(

Using left endpoints

=xnx1 x2xi-1 xi

x

y

a

y = f(x)

bx0=

=xnx1 x2 xi-1xi

x

y

a

y = f(x)

bx0=

f(xi-1)

∆x

f(xi-1)

Area of i-th column is f(xi-1)•∆x

=xnx1 x2 xi-1xi

x

y

a

y = f(x)

bx0=

f(xi-1)

∆x

Total area of all columns is

xxfxxfxxfxxf ni )(...)(...)()( 1110

xxfLn

iin

11)(

=xnx1 x2 xi-1xi

x

y

a bx0=

Note: Ln ≠ Rn

Ln - Rn = ?

xxfxxfxxfxxfR nin )(...)(...)()( 21

xxfxxfxxfxxfL nin )(...)(...)()( 120

Nevertheless…

• Theorem. If f(x) is continuous on [a,b], then both limits and

exist and

nnR

lim n

nL

lim

nn

nn

LR

limlim

x*nx*ix*2

Using sample points

=xnx1 x2xi-1 xi

x

y

a

y = f(x)

bx0=

Choose a sample point - an arbitrary point xi* in [xi-1, xi]for each i

x*1

xi-1x1 x2

Area of i-th column is f(xi*)•∆x

x

y

a

y = f(x)

bx0= =xnxi

f(xi*)

∆x

f(xi*)

x*i

f(xi*)∆x

Total area of all columns is

xxfxxfxxfxxf ni )(...)(...)()( ***2

*1

xxfn

ii

1

*)(

x*i=xnx1 x2

xi-1 xi

x

y

a

y = f(x)

bx0= x*1 x*2 x*n

Theorem

• If f(x) is continuous on [a,b], then the limit

exists and does not depend on the choice of sample points

xxfn

ii

n

1

*)(lim

4.2 Definite Integral

• Now we consider functions that may change sign on [a,b]

• In this case, we need to take into account sign of f(x)

• Idea: use “signed area”

A3

A2

A1

Signed area

x

y

a

y = f(x)

b

“Net Area” = A1 – A2 + A3

x*n

x*i

x*2

Choose sample points

=xnx1 x2

xi-1 xi x

y

a

y = f(x)

bx0= x*1

x*i

Signed area of i-th column is f(xi*)•∆x

=xnx1 x2

xi-1 xi x

y

a

y = f(x)

bx0=

f(xi*)

∆x

Net area of all columns is

xxfxxfxxfxxf ni )(...)(...)()( ***2

*1

xxfn

ii

1

*)(

x*n

x*i

x*2=xnx1 x2

xi-1 xi x

y

a

y = f(x)

bx0= x*1

Riemann Sum that correspond ton and given choice of sample points

xxfn

ii

1

*)(

x*n

x*i

x*2=xnx1 x2

xi-1 xi x

y

a

y = f(x)

bx0= x*1

Definite Integral of function f from a to b is defined as the limit of Riemann sums

xxfdxxfn

ii

b

an

1

*)(lim)(

x*n

x*i

x*2=xnx1 x2

xi-1 xi x

y

a

y = f(x)

bx0= x*1

Theorem

• If f(x) is continuous on [a,b], then the definite integral of function f from a to b

exists and does not depend on the choice of sample points

xxfdxxfn

ii

b

an

1

*)(lim)(

Terminology

b

a

dxxf )(Lower limit

Upper limit

Integral sign

Integrand

Definite Integral in terms of area:

b

a

AAAdxxf 321)(

A3

A2

A1

x

y

a

y = f(x)

b