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INTEGRALS 5

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5.1 Areas and Distances INTEGRALS In this section, we will learn that: We get the same special type of limit in trying to find the area under a curve or a distance traveled.

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AREA PROBLEM We begin by attempting to solve the area problem: Find the area of the region S that lies under the curve y = f(x) from a to b.

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AREA PROBLEM This means that S, illustrated here, is bounded by: The graph of a continuous function f [where f(x) ≥ 0] The vertical lines x = a and x = b The x-axis

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RECTANGLES For a rectangle, the area is defined as: The product of the length and the width

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TRIANGLES The area of a triangle is: Half the base times the height

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POLYGONS The area of a polygon is found by: Dividing it into triangles and adding the areas of the triangles

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AREA PROBLEM However, it isn’t so easy to find the area of a region with curved sides. We all have an intuitive idea of what the area of a region is. Part of the area problem, though, is to make this intuitive idea precise by giving an exact definition of area.

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AREA PROBLEM Recall that, in defining a tangent, we first approximated the slope of the tangent line by slopes of secant lines and then we took the limit of these approximations. We pursue a similar idea for areas.

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AREA PROBLEM We first approximate the region S by rectangles and then we take the limit of the areas of these rectangles as we increase the number of rectangles. The following example illustrates the procedure.

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AREA PROBLEM Use rectangles to estimate the area under the parabola y = x 2 from 0 to 1, the parabolic region S illustrated here. Example 1

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AREA PROBLEM We first notice that the area of S must be somewhere between 0 and 1, because S is contained in a square with side length 1. However, we can certainly do better than that. Example 1

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AREA PROBLEM Suppose we divide S into four strips S 1, S 2, S 3, and S 4 by drawing the vertical lines x = ¼, x = ½, and x = ¾. Example 1

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AREA PROBLEM We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip. Example 1

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AREA PROBLEM In other words, the heights of these rectangles are the values of the function f(x) = x 2 at the right endpoints of the subintervals [0, ¼],[¼, ½], [½, ¾], and [¾, 1]. Example 1

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AREA PROBLEM Each rectangle has width ¼ and the heights are (¼) 2, (½) 2, (¾) 2, and 1 2. Example 1

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AREA PROBLEM If we let R 4 be the sum of the areas of these approximating rectangles, we get: Example 1

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AREA PROBLEM We see the area A of S is less than R 4. So, A < 0.46875 Example 1

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AREA PROBLEM Instead of using the rectangles in this figure, we could use the smaller rectangles in the next figure. Example 1

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AREA PROBLEM Here, the heights are the values of f at the left endpoints of the subintervals. The leftmost rectangle has collapsed because its height is 0. Example 1

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AREA PROBLEM The sum of the areas of these approximating rectangles is: Example 1

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AREA PROBLEM We see the area of S is larger than L 4. So, we have lower and upper estimates for A: 0.21875 < A <0.46875 Example 1

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AREA PROBLEM We can repeat this procedure with a larger number of strips. Example 1

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AREA PROBLEM The figure shows what happens when we divide the region S into eight strips of equal width. Example 1

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AREA PROBLEM By computing the sum of the areas of the smaller rectangles (L 8 ) and the sum of the areas of the larger rectangles (R 8 ), we obtain better lower and upper estimates for A: 0.2734375 < A < 0.3984375 Example 1

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AREA PROBLEM So, one possible answer to the question is to say that: The true area of S lies somewhere between 0.2734375 and 0.3984375 Example 1

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AREA PROBLEM We could obtain better estimates by increasing the number of strips. Example 1

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AREA PROBLEM The table shows the results of similar calculations (with a computer) using n rectangles, whose heights are found with left endpoints (L n ) or right endpoints (R n ). Example 1

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AREA PROBLEM From the values in the table, it looks as if R n is approaching 1/3 as n increases. We confirm this in the next example.

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AREA PROBLEM For the region S in Example 1, show that the sum of the areas of the upper approximating rectangles approaches 1/3, that is, Example 2

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AREA PROBLEM R n is the sum of the areas of the n rectangles. Each rectangle has width 1/n and the heights are the values of the function f(x) = x 2 at the points 1/n, 2/n, 3/n, …, n/n. That is, the heights are (1/n) 2, (2/n) 2, (3/n) 2, …, (n/n) 2. Example 2

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AREA PROBLEM Thus, Example 2

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AREA PROBLEM Here, we need the formula for the sum of the squares of the first n positive integers: Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E. E. g. 2—Formula 1

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AREA PROBLEM Putting Formula 1 into our expression for R n, we get: Example 2 If n is near Infinity, on can see that the above is

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AREA PROBLEM It can be shown that the lower approximating sums also approach 1/3, that is,

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AREA PROBLEM Thus, we define the area A to be the limit of the sums of the areas of the approximating rectangles, that is,

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AREA PROBLEM We start by subdividing S into n strips S 1, S 2, …., S n of equal width.

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AREA PROBLEM Let’s apply the idea of Examples 1 and 2 to the more general region S of the earlier figure.

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AREA PROBLEM The width of the interval [a, b] is b – a. So, the width of each of the n strips is:

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AREA PROBLEM These strips divide the interval [a, b] into n subintervals [x 0, x 1 ], [x 1, x 2 ], [x 2, x 3 ],..., [x n-1, x n ] where x 0 = a and x n = b.

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AREA PROBLEM The right endpoints of the subintervals are: x 1 = a + ∆x, x 2 = a + 2 ∆x, x 3 = a + 3 ∆x,.

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AREA PROBLEM Let’s approximate the i th strip S i by a rectangle with width ∆x and height f(x i ), which is the value of f at the right endpoint. Then, the area of the i th rectangle is f(x i )∆x.

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AREA PROBLEM What we think of intuitively as the area of S is approximated by the sum of the areas of these rectangles: R n = f(x 1 ) ∆x + f(x 2 ) ∆x + … + f(x n ) ∆x

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AREA PROBLEM Here, we show this approximation for n = 2, 4, 8, and 12.

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AREA PROBLEM Notice that this approximation appears to become better and better as the number of strips increases, that is, as n → ∞.

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AREA PROBLEM Therefore, we define the area A of the region S as follows.

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AREA PROBLEM It can be proved that the limit in Definition 2 always exists—since we are assuming that f is continuous.

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AREA PROBLEM It can also be shown that we get the same value if we use left endpoints: Equation 3

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SAMPLE POINTS In fact, instead of using left endpoints or right endpoints, we could take the height of the i th rectangle to be the value of f at any number x i * in the i th subinterval [x i - 1, x i ]. We call the numbers x i *, x 2 *,..., x n * the sample points.

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AREA PROBLEM The figure shows approximating rectangles when the sample points are not chosen to be endpoints.

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AREA PROBLEM Thus, a more general expression for the area of S is: Equation 4

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SIGMA NOTATION We often use sigma notation to write sums with many terms more compactly. For instance,

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AREA PROBLEM Hence, the expressions for area in Equations 2, 3, and 4 can be written as follows: Later on we set the limit of the sum above to be what we call definite integral

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DISTANCE PROBLEM Now, let’s consider the distance problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. In a sense, this is the inverse problem of the velocity problem that we discussed in Section 2.1

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CONSTANT VELOCITY If the velocity remains constant, then the distance problem is easy to solve by means of the formula distance = velocity x time

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VARYING VELOCITY However, if the velocity varies, it’s not so easy to find the distance traveled. We investigate the problem in the following example.

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DISTANCE PROBLEM In general, suppose an object moves with velocity v = f(t) where a ≤ t ≤ b and f(t) ≥ 0. So, the object always moves in the positive direction.

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DISTANCE PROBLEM The distance d traveled is indeed the limit of : Later, as we said before, we will find tha Equation 5

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SUMMARY Equation 5 has the same form as our expressions for area in Equations 2 and 3. So, it follows that the distance traveled is equal to the area under the graph of the velocity function.

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SUMMARY So, when we compute areas in this chapter, bear in mind that they can be interpreted in a variety of practical ways.

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SUMMARY In Chapters 6 and 9, we will see that other quantities of interest in the natural and social sciences can also be interpreted as the area under a curve. Examples include: Work done by a variable force Cardiac output of the heart

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