10 June 2007 Prepared By: Fatma Abdulrab AlJazi Nassre Rasha Fatma Al-Malki Reem Saleh & Safa’a Al-aamri

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.

10 June 2007 “God does not care about our mathematical difficulties - he integrates empirically” ALBERT EINSTEIN

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:

10 June 2007 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.

10 June 2007 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.

10 June 2007 From the previous illustration of dividing the area under a curve to rectangles or trapeziaMathematicians reached the following conclusion: From the previous illustration of dividing the area under a curve to rectangles or trapezia Mathematicians reached the following conclusion: ( 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 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

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

10 June 2007 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:

10 June 2007 Example 2:

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Area of both shaded areas is = ( ¼ x 2 ) = 2 units squared. = 2 units squared.

10 June 2007 Let us find the area of the region bounded by the graphs of f ( x ) = x 2 and g ( x ) = 8 f ( x ) = x 2 and g ( x ) = 8 First let us graph these functions. Notice that the region is bounded above by g(x) and below by f (x). To find the boundary points which will give the vertical side lines we need to solve f ( x ) = g ( x ) or x 2 = 8 f ( x ) = g ( x ) or x 2 = 8 Easily we get x = 0 and x = 4. So the answer is Area ( ) = (8 - x 2 ) dx We have 8 - x 2 dx = x 3/2 - x 3 =.

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Significance 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 = x 2 from x = 0 to x = 2:

10 June 2007 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: where A(x( is the area the cross section at a point x.

10 June 2007 Revolving a plane figure about an axis generates a volume Definition: Consider the region between the graph of a continuous function y = f(x) and the x-axis from x = a to x = b.

10 June 2007 A Famous Paradox A 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 rotating an area about the y-axis can be found in a way similar to that about the x-axis. The volume of such solid of revolution is given by Remember that dy implies that a and b are y limits.

10 June 2007 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 about the y-axis. The required volume is given by: Now So

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