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1.1 Limits: an Informal View 4.1 Areas and Sums Chapter 4 Integrals from Calculus: Concepts & Computation by David Betounes and Mylan Redfern | 2016 Copyright | 9781524917692 Property of Kendall Hunt Publishing 4.1 Areas and Sums 257 4 Integrals Calculus Concepts & Computation 1.1 Limits: An Informal View 4.1 Areas and Sums > with(code_ch4,circle,parabola);with(code_ch1sec1); [circle, parabola] [polygonlimit, secantlimit] This chapter begins our study of the integral calculus, which is the counter-part to the differential calculus you learned in Chapters 1–3. The topic of integrals, otherwise known as antiderivatives, was introduced in Section 3.7, and you should The General Area Problem read that discussion before continuing here. A major application of antiderivatives, or integrals, is the measurement of areas of planar regions D. In fact, a principal motive behind the invention of the integral calculus was to solve the area problem. See Figure 4.1.1. D Before we learn how to measure areas of such general regions D, it is best to think about the areas of simple regions, and, indeed, reflect on the very notion of what is meant by “area.” This we discuss here. Then, in the next two sections, Sections 4.2 and 4.3 , we look at the details of measuring areas of regions “beneath the graphs of continuous functions.” Then in Section 5.1 we tackle the task of finding area of more complicated regions, such as the region D in Figure 4.4.1. We have intuitive notions about what is meant by the area of a region. It is a positive Find the area Figure 4.1.1: number A which somehow measures the “largeness” of the region and enables us of the region D bounded by a to compare regions with different shapes. Knowing the formula for the area of a piecewise smooth curve . rectangle allows us to find the areas of other basic regions, such as parallelograms, triangles, and trapezoids—all of which can be found without calculus. If, however, we want to find areas of planar regions with curved boundaries, such as the region D in Figure 4.1.1, then we need to use calculus. Areas of Basic Regions You probably “know” that the area of a rectangle is the product of its base times its height A = bh. But you should reflect on the fact that this is actually a definition, rather than something that can be derived from more primitive notions. DEFINITION 4.1.1 (Area of a Rectangle) The area of a rectangle with base b and height h is A = bh. h b It is said that in ancient times, the measurement of the length of the boundary of a region (its perimeter) was used sometimes to quantify the largeness of the region. After all, length is easier to measure—you just step off along the boundary, counting the paces as you go. But for quantifying the extent of a rectangular piece of property, the area A = bh, given as the product of the lengths of its sides, turned out to be a better gauge of the value of the property. Chapter 4 Integrals from Calculus: Concepts & Computation by David Betounes and Mylan Redfern | 2016 Copyright | 9781524917692 Property of Kendall Hunt Publishing 258 Chapter 4: Integrals Using this as the definition of the area of a rectangle, we can derive the formulas for the areas of other basic regions. For this, we need to use the primitive notions: Q ¢ R (1) The areas of congruent figures are the same. h (2) If a region is the disjoint union of subregions, then the area of the region is the sum of the areas of the subregions. P Q Applying these primitive notions to the case of a right triangle, we get that its area b is one-half its base times its height. You can easily prove this as an exercise. Next, how do we deal with parallelograms that are not rectangles and triangles Figure 4.1.2: The area of bh 1 that are not right triangles? Why is it that the formulas for their areas are A = a right triangle is A = bh. and A = 1 bh, respectively? To see this, first consider the case of a parallelogram. 2 2 DEFINITION: A parallelogram is a quadrilateral (four-sided figure) whose opposite sides are parallel (and of equal length). Figure 4.1.3 shows a rectangle PQRS and parallelogram P QR0S0 which have the same base b and height h. S R S ¢ R¢ A3 A1 h A4 A2 P b Q Figure 4.1.3: The area of a parallelogram is also A = bh. To show that the rectangle and parallelogram have the same area, label the areas of four of subregions as A1,A2,A3,A4 as shown. By elementary geometry one can prove that the right triangles P SS0 and QRR0 are congruent (exercise), and hence they have the same area. So A1 + A3 = A3 + A4. Hence, A1 = A4. Adding A2 to both sides of this gives A1 + A2 = A2 + A4. But A1 +A2 is the area of rectangle PQRS and A2 +A4 is the area of parallelogram b P QR0S0. Thus, the two areas are equal. The measure of each is A = bh, since that S 2 R is the measure for the area of the rectangle. This result for areas of parallelograms leads to the area formula for general h triangles (not just right triangles). This is left as an exercise. Using these formulas x y for the areas of basic regions, we can find formulas for areas of more complicated P M N Q regions. b 1 Example 4.1.1 (The Trapezoidal Area Formula) Figure 4.1.4: A trapezoid Problem: Using area formulas for triangles and rectangles, derive the formula for the with bases b1, b2 and height h area of the trapezoid given in Figure 4.1.4. (Recall that a trapezoid is a quadrilateral 1 has area A = 2 (b1 + b2) h. with one pair of opposite sides that are parallel.) The formula is 1 A = (b + b ) h. 2 1 2 Chapter 4 Integrals from Calculus: Concepts & Computation by David Betounes and Mylan Redfern | 2016 Copyright | 9781524917692 Property of Kendall Hunt Publishing 4.1 Areas and Sums 259 This says that the area is the average of the bases b1, b2 times the height h. Also note that when the bases have the same length, b1 = b2 = b, the trapezoid is a parallelogram and the formula reduces to A = bh. Solution: Dropping perpendiculars from points R and S onto the base PQ gives points M and N. This also divides the trapezoid into a rectangle and two right trian- gles. Labeling the bases PM and QN of the right triangles by x and y, respectively, gives the areas of these triangles as 1 1 A = xh, A = yh 1 2 2 2 The area of the rectangle MNRS is A3 = b2h. It is clear that b1 = x + b2 + y, or, equivalently x + y = b b . 1 − 2 Now use this last expression in the following calculation of the area of the trapezoid. 1 1 A = A + A + A = xh + yh + b h 1 2 3 2 2 2 1 1 = (x + y) + b h = (b b ) + b h 2 2 2 1 − 2 2 1 = (b + b ) h. 2 1 2 This gives the formula that we were looking for. 3 Example 4.1.2 Problem: Apply the trapezoid formula from the previous example to find the area 2 of the particular trapezoid in Figure 4.1.5. Solution: Here the bases are b1 = 5, b2 = 3 ,and the height is h = 2. So the 5 computation is easy enough: Figure 4.1.5: A trapezoid. 1 A = (5 + 3) 2 = 8. 2 Example 4.1.3 (Five Adjacent Trapezoids) Problem: Determine the formula for the area of a region made of 5 trapezoids as shown in Figure 4.1.6. b6 h b5 h b4 h b3 h b2 h b1 Figure 4.1.6: A region composed of 5 trapezoids Chapter 4 Integrals from Calculus: Concepts & Computation by David Betounes and Mylan Redfern | 2016 Copyright | 9781524917692 Property of Kendall Hunt Publishing 260 Chapter 4: Integrals Solution: The trapezoids in Figure 4.1.6 all have the same height h and have bases taken in pairs form the sides labeled b1, b2, . , b6 . Using the formula for the area of a trapezoid, derived in Example 4.1.1, and the additivity principle for areas in general, we find the area of the region in Figure 4.1.6 is 1 1 1 1 1 A = (b + b ) h + (b + b ) h + (b + b ) h + (b + b ) h + (b + b ) h 2 1 2 2 2 3 2 3 4 2 4 5 2 5 6 1 = (b + b + b + b + b + b + b + b + b + b ) h 2 1 2 2 3 3 4 4 5 5 6 1 = (b + 2 b + 2 b + 2 b + 2 b + b ) h. 2 1 2 3 4 5 6 This is sum of the averages of all the bases (counting the overlapping ones on the interior twice) times the common height h. Note that the formula would not be so simple if the heights of the trapezoids were different. Areas of Regions Under Graphs Later we will be interested in finding areas of regions bounded by graphs of func- y4 tions. For step-functions and polygonal functions we can use the ideas of the pre- y 1 y vious examples to compute the area. Before doing this note that if x1 < x2 are y 3 2 points on the x-axis, then we use y ∆x = x2 x1 5 − to denote the length of the line segment between these points.
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