Integration Formula

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  253 5 In this chapter we will introduce “integration,” a process motivated by the problem of computing the area of plane regions. After an informal overview of the problem, we will discuss a surprising relationship between integration and differentiation that is known as the Fundamental Theorem of Calculus. We will then apply integration to continue our study of rectilinear motion and to define the “average value" of a function. We conclude the chapter by studying functions that can be defined using integration, with a focus on the natural logarithm function. INTEGRATION 5.1 AN OVERVIEW OF THE AREA PROBLEM In this introductory section we will consider the problem of calculating areas of plane regions with curvilinear boundaries. All of the results in this section will be reexamined in more detail later in this chapter. Our purpose here is simply to introduce and motivate the fundamental concepts. THE AREA PROBLEM Formulas for the areas of polygons, such as squares, rectangles, triangles, and trapezoids, were well known in many early civilizations. However, the problem of finding formulas for regions with curved boundaries (a circle being the simplest example) caused difficulties for early mathematicians. The first real progress in dealing with the general area problem was made by the Greek mathematician Archimedes, who obtained areas of regions bounded by circular arcs, parabolas, spirals, and various other curves using an ingenious procedure that was later called the method of exhaustion. The method, when applied to a circle, consists of inscrib- ing a succession of regular polygons in the circle and allowing the number of sides to in- crease indefinitely (Figure 5.1.1). As the number of sides increases, the polygons tend to “exhaust” the region inside the circle, and the areas of the polygons become better and better approximations of the exact area of the circle.

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  Calculus: An Applied Approach, Brief, International Metric Edition

  • Ron Larson(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Evaluating a Definite Integral Using a Geometric Formula The definite integral integral.alt1 2 0 2 x d x represents the area of the region bounded by the graph of f ( x ) = 2 x , the x -axis, and the line x = 2, as shown in Figure 5.5. The region is triangular, with a height of 4 units and a base of 2 units. Using the formula for the area of a triangle, you have integral.alt1 2 0 2 x dx = 1 2 ( base )( height ) = 1 2 ( 2 )( 4 ) = 4 . Checkpoint 1 Worked-out solution available at LarsonAppliedCalculus.com Evaluate the definite integral using a geometric formula. Illustrate your answer with an appropriate sketch. integral.alt1 3 0 4 x d x b a x R y y = f ( x ) integral.alt1 b a f ( x ) dx = Area In Exercise 79 on page 350, you will use integration to find a model for the mortgage debt outstanding for one- to four-family homes. 4 3 2 1 4 2 1 3 x y f ( x ) = 2 x FIGURE 5.5 Originalpunkt/Shutterstock.com Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 340 Chapter 5 Integration and Its Applications The Fundamental Theorem of Calculus Consider the function A , which denotes the area of the region shown in Figure 5.6. x b x a y y = f ( x ) A ( x ) = Area from a to x FIGURE 5.6 To discover the relationship between A and f , let x increase by an amount Δ x . This increases the area by Δ A . Let f ( m ) and f ( M ) denote the minimum and maximum values of f on the interval [ x , x + Δ x ] . x x x b x a b x a b x a y y y x + uni0394 x x + uni0394 x x + uni0394 x f ( m ) uni0394 x f ( M ) uni0394 x uni0394 A f ( m ) f ( M ) FIGURE 5.7 As indicated in Figure 5.7, you can write the inequality below. f ( m ) Δ x ≤ Δ A ≤ f ( M ) Δ x See Figure 5.7. f ( m ) ≤ Δ A Δ x ≤ f ( M ) Divide each term by Δ x . lim Δ x uni2192 0 f ( m ) ≤ lim Δ x uni2192 0 Δ A Δ x ≤ lim Δ x uni2192 0 f ( M ) Take limit of each term.

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  Calculus

  Early Transcendental Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  The open form indicates the summands and the closed form is an explicit formula for the sum. TECHNOLOGY MASTERY If you have access to a CAS, it will provide a method for finding closed forms such as those in Theorem 5.4.2. Use your CAS to confirm the formulas in that theorem, and then find closed forms for n  k=1 k 4 and n  k=1 k 5 Example 3 Express n  k=1 (3 + k) 2 in closed form. Solution. n  k=1 (3 + k) 2 = 4 2 + 5 2 + · · · + (3 + n) 2 = [1 2 + 2 2 + 3 2 + 4 2 + 5 2 + · · · + (3 + n) 2 ] − [1 2 + 2 2 + 3 2 ] =  3+n  k=1 k 2  − 14 = (3 + n)(4 + n)(7 + 2n) 6 − 14 = 1 6 (73n + 21n 2 + 2n 3 ) A DEFINITION OF AREA We now turn to the problem of giving a precise definition of what is meant by the “area under a curve.” Specifically, suppose that the function f is continuous and nonnegative on the interval [a, b], and let R denote the region bounded below by the x-axis, bounded on the sides by the vertical lines x = a and x = b, and bounded above by the curve y = f (x) (Figure 5.4.2). Using the rectangle method of Section 5.1, we can motivate a definition for Figure 5.4.2 the area of R as follows: 278 Chapter 5 / Integration • Divide the interval [a, b] into n equal subintervals of width Δx = b − a n by inserting n − 1 equally spaced points x 1 , x 2 , . . . , x n−1 (Figure 5.4.3). Figure 5.4.3 • Over each subinterval construct a rectangle whose height is the value of f at a se- lected point in the subinterval. Thus if x ∗ 1 , x ∗ 2 , . . . , x ∗ n denote the n selected points, then the rectangles will have heights f (x ∗ 1 ), f (x ∗ 2 ), . . . , f (x ∗ n ) and areas f (x ∗ 1 )Δx, f (x ∗ 2 )Δx, . . . , f (x ∗ n )Δx (Figure 5.4.4). • The union of the rectangles forms a region R n whose area can be regarded as an approximation to the area A of R; that is, A = area(R) ≈ area(R n ) = f (x ∗ 1 )Δx + f (x ∗ 2 )Δx + · · · + f (x ∗ n )Δx = n  k=1 f (x ∗ k )Δx (Figure 5.4.5).

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