10 June 2007

Prepared By:

Fatma AbdulrabAlJazi Nassre

RashaFatma Al-Malki

Reem Saleh

&

Safa’a Al-aamri

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1. What is Integration all about.

2. Using Integration to find the Area under a curve.

3. Using Integration to find the Volume of Revolution.

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““God does not care about our mathematical difficulties - he integrates empirically”

ALBERT EINSTEIN

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We Know that if

Know suppose that we are given and asked to find y in terms of x.

This process is the reverse of differentiation and is called:

Integration.

In a particular case, we know that

But so will

In fact, , where c is a constant,

will also satisfy

For this reason, we write ,where c is called the constant of integration.

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Example 1:

Find:

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We already know that :

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Example 3:

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Integrals and Area Integrals and Area

It is an odd fact that the basic ideas of integration predate those of differentiation by nearly 2000 years. Archimedes was quite good at `doing integrals', albeit in a rather different context. He was interested in the problem of calculating the areas of curved regions. The ancient Egyptians and Babylonians were quite good at finding the areas of straight-sided regions (basic surveying--much needed in Egypt because of the regular flooding of the Nile). Curved regions are an altogether different problem and the ancient Ancients knew little more than that the area of a circle is r 2. More complicated regions need much more complicated ideas. It is not enough just to know the area of a triangle.

Archimedes, in an remarkable anticipation of the methods of the Calculus, tried to adopt a `limiting values' approach to calculating areas. His idea was that you could approximate a curved region arbitrarily closely by straight-sided regions and could therefore hope to obtain the area of the curved region as a limiting case of the (calculable) areas of the approximations. This has since become one of the most basic ways of defining integrals because, as we will see in this section, there is a close connection between the idea of an integral (which we just introduced as the opposite of differentiation) and the idea of an area.

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Our first need is for a notation with which to describe an area bounded by curves. A good start will be to consider the area between the graph of a function and the x-axis.

We could simply count the squares under the curve using graph paper, or we could split the area into thin strips: rectangles or trapezia which would give a better estimation of the area under the curve.

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From the previous illustration of dividing the area under a curve to From the previous illustration of dividing the area under a curve to rectangles or trapeziarectangles or trapezia Mathematicians reached the following Mathematicians reached the following conclusion: conclusion:

( if you would like to know more information of how Mathematicians reached the following ( if you would like to know more information of how Mathematicians reached the following formulas for finding the Area under a curve ,contact any of the members to send you the formulas for finding the Area under a curve ,contact any of the members to send you the complete report ).complete report ).

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If 0 < f(x) < g(x) for all x in [a, b], then

If a < b then it is convenient to define

11..

22..

33..

44..

55..

66..

77..

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Between Curve & x- axisBetween Curve & x- axis Between Curve & y- axisBetween Curve & y- axis

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And since we all know that area can never be negative since it’s a kind of measurement, we can add a negative sign at the beginning, before the integral sign, so that it cancels the one resulted in the answer of the integration and which represents the Area.

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Example 1:

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Example 2:Example 2:

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Area of both shaded areas is = ( ¼ Area of both shaded areas is = ( ¼ xx 2 ) 2 )

= 2 units squared. = 2 units squared.

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Let us find the area of the region Let us find the area of the region     bounded by the graphs of   bounded by the graphs of

ff ( (xx) = ) = xx22    and        and    gg((xx) = 8) = 8           

First let us graph these functions. First let us graph these functions.

Notice that the region is bounded above by Notice that the region is bounded above by gg((xx) and below by ) and below by ff ( (xx).).To find the boundary points which will give the To find the boundary points which will give the vertical side lines we need to solve vertical side lines we need to solve

ff ( (xx) = ) = gg((xx)    or      )    or      xx22 = 8 = 8          

Easily we get Easily we get xx = 0 and = 0 and xx = 4. So the answer is = 4. So the answer is Area(   ) =        (8      - x2 )   dx  

We haveWe have

             8           - x2     dx =            x3/2 -     x3        =          .

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Significance of Volumes and SurfacesSignificance of Volumes and Surfaces

The definite integral is an amazingly versatile tool. In the coming Examples we will see how a rotated plane figure sweeps out a volume.

But the process of visualizing this one concept has far wider applications.

We can compute area, volume, arc length and surface area using essentially the same mental process. First we divide an object into smaller pieces -  n smaller pieces of a thickness that will eventually become our dx or dy.  We approximate a quantity for each of the small pieces.  This is usually an area or a length.  We add up the approximations and then take a limit.  Thus, we have intuitively derived a definite integral.

Sketch the solid and a typical cross section. Find a formula for the crioss-sectional area A(x). Find the limits for integration on the rotational axis. Integrate A(x) to find the volume.

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Consider the solid of revolution formed by the graph of y = x2 from x = 0 to x = 2:

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Suppose you wanted to make a clay vase.  It is made by shaping the clay into a curve and spinning it along an axis.  If we want to determine how much water it will hold, we can consider the cross sections that are perpendicular to the axis of rotation, and add up all the volumes of the small cross sections.  We have the following definition:

wherewhere AA((xx(( is the area the cross section at a point is the area the cross section at a point xx..

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Revolving a plane figure about an axis generates a volume Revolving a plane figure about an axis generates a volume

DefinitionDefinition::  Consider the region between the graph of a continuous   Consider the region between the graph of a continuous function  function  yy = = ff((xx) and the ) and the xx-axis from  -axis from  xx =  =  aa to to xx =  =  b.b.

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A Famous ParadoxA Famous Paradox   Gabriel's Horn or Torricelli's Trumpet

                                      

If the function  y = 1/x   is revolved around the x-axis for x > 1,the figure has a finite volume, but infinite surface area.

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The volume of solid of revolution formed by The volume of solid of revolution formed by rotating an area about the y-axis can be found rotating an area about the y-axis can be found in a way similar to that about the x-axis.in a way similar to that about the x-axis.

The volume of such solid of revolution is given The volume of such solid of revolution is given byby

Remember that dy implies that a and b are y Remember that dy implies that a and b are y limits.limits.

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Find the volume of the solid formed when the area between Find the volume of the solid formed when the area between the curve and the y-axis from y=1 to y=8 is rotated the curve and the y-axis from y=1 to y=8 is rotated about the y-axis.about the y-axis.

The required volume is given by:

NowSoSo

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