Applications of the Definite Integrals Dr. Faud Almuhannadi Math 119 - Section(4)

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Transcript of Applications of the Definite Integrals Dr. Faud Almuhannadi Math 119 - Section(4)

Applications of the Definite Integrals

Dr. Faud AlmuhannadiMath 119 - Section(4)

Done by:

Hanen Marwa Najla Noof Wala

In this part, we are going to explain the

different types of applications related to

the “ Definite Integrals “.

Which includes talking about :

1. Area under a curve

2. Area between two curves

3. Volume of Revolution

Definition :

In calculus, the integral of a function

extends the concept of an ordinary

sum. While an ordinary sum is taken

over a discrete set of values,

integration extends this concept to

sums over continuous domains

The simplest case, the integral of a real-

valued function f of one real variable x on

the interval [a, b], is denoted:

The ∫ sign represents integration; a and b

are the lower limit and upper limit of

integration, defining the domain of

integration; f(x) is the integrand; and dx is

a notation for the variable of integration

Computing integrals

The most basic technique for

computing integrals of one real

variable is based on the fundamental

theorem of calculus. It proceeds like

this:

Choose a function f(x) and an interval [a, b].

Find an antiderivative of f, that is, a function F such that F' = f.

By the fundamental theorem of calculus, provided the integrand and integral have no singularities on the path of integration,

Therefore the value of the integral is F(b) − F(a).

Case ..1..

Area Under a Curve

Example ..1..

The graph below shows the curve and is shaded in the region

The area is found by integrating

Example ..2..

Case ..2..

Area between two curves

Say you have 2 curves y = f(x) and y = g(x)

Area under f(x)=

Area under g(x)=

Superimposing the two graphs:

Area bounded by f(x) and g(x)

Example ..3..

 Find the area between the curves        y = 0      and      y = 3(x3 - x)

1 2 3 4-1-2-3-4

x

1

2

3

4

-1

-2

-3

-4

y

Example ..4.. Find the area bounded by the curves

y = x2 - 4x – 5

and

y = x + 1

Solving the equations simultaneously,

          x + 1 = x2 - 4x - 5

           x = -1 or x = 6

Required Area =

Volume Of A Revolution

A solid of revolution is formed when a region bounded by part of a curve is rotated about a straight line.

  Rotation about x-axis:

Rotation about y-axis:

Example ..5..

The volume that we are looking for is shown in the diagram below

To find the volume, we integrate

Thank u 4 listening