advertisement

Chapter 13 – 4 The Definite Integral Approximating Areas Using Left and Right Sums y = 2x 12 10 Example 1. Find the approximate area between the graph of y = 2x and the xaxis from x = 2 to x = 4. (The area is 12.) 8 6 4 2 We can approximate the area by finding the sum of the areas of vertical rectangles where each rectangle has the same base. The altitude of each rectangle can be the function value calculated for the x-coordinate of the left end of its base, the right end of its base, or the middle of its base (respectively): y = 2x 12 10 10 10 8 8 8 6 6 6 4 4 4 2 2 2 0 0 2 3 4 5 1 2 3 4 5 y = 2x 12 1 0 y = 2x 12 0 0 0 0 1 2 3 4 1 1 1 1 1 2 2 2 2 2 5 0 1 2 3 4 5 Left Sum = 4 ∙ + 5 ∙ + 6 ∙ + 7 ∙ = (4 + 5 + 6 + 7) ∙ = 11 1 1 1 1 1 2 2 2 2 2 Right Sum = 5 ∙ + 6 ∙ + 7 ∙ + 8 ∙ = (5 + 6 + 7 + 8) ∙ = 13 1 1 1 1 1 2 2 2 2 2 Midpoint Sum = 4.5 ∙ + 5.5 ∙ + 6.5 ∙ + 7.5 ∙ = (4.5 + 5.5 + 6.5 + 7.5) ∙ = 12 To improve the accuracy of the approximation, increase the number of rectangles. Suppose we use 8 rectangles instead of 4: y = 2x y = 2x y = 2x 12 12 12 10 10 10 8 8 8 6 6 6 4 4 4 2 2 2 0 0 0 1 2 3 4 5 0 0 1 2 3 4 5 0 1 Left Sum = (4 + 4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5) ∙ = 11.5 4 1 Right Sum = (4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5 + 8) ∙ = 12.5 4 1 Midpoint Sum = (4.25 + 4.75 + 5.25 + 5.75 + 6.25 + 6.75 + 7.25 + 7.75) ∙ = 12 4 1 2 3 4 5 Example 2. Find the approximate area between the graph of 𝑦 = √4𝑥 − 𝑥 2 and the x-axis from x = 0 to x = 2. The 1 region is a semi-circle and the actual area is 𝜋𝑟 2 = 2𝜋 ≈ 6.283. 2 As we did above, we will find the sum of the areas of vertical rectangles to approximate the area under the curve: y = sqrt(4x - x2) y = sqrt(4x - x2) y = sqrt(4x - x2) 2 2 2 1 1 1 0 0 0 1 2 3 4 0 0 1 2 3 4 0 1 2 3 4 3 4 Left Sum ≈ (0 + 1.73 + 2 + 1.73) ∙ 1 = 5.46 Right Sum ≈ (1.73 + 2 + 1.73 + 0) ∙ 1 = 5.46 Midpoint Sum = (1.32 + 1.94 + 1.94 + 1.32) ∙ 1 = 6.52 Doubling the number of rectangles to 8 gives us a better approximation: y = sqrt(4x - x2) y = sqrt(4x - x2) y = sqrt(4x - x2) 2 2 2 1 1 1 0 0 0 1 2 3 4 0 0 1 2 3 4 0 1 1 Left Sum ≈ (0 + 1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32) ∙ = 5.99 2 1 Right Sum ≈ (1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32 + 0) ∙ = 5.99 2 1 Midpoint Sum ≈ (0.97 + 1.56 + 1.85 + 1.98 + 1.98 + 1.85 + 1.56 + 0.97) ∙ = 6.36 2 2 Riemann Sum The left sum, the right sum and the midpoint sum are all special instances of a Riemann Sum. Assume that a function f(x) is continuous on the interval [a,b]. Divide the interval [a,b] into n sub-intervals of width∆𝑥 = 𝑏−𝑎 𝑛 . These subintervals have endpoints 𝑥0 , 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 where 𝑥0 = 𝑎, 𝑥𝑛 = 𝑏, and 𝑥𝑖 − 𝑥𝑖−1 = ∆𝑥 for any value of i from 1 to n. (This implies that 𝑥𝑖 = 𝑎 + 𝑖∆𝑥 for any i from 1 to n.) If 𝑥𝑖−1 ≤ 𝑐𝑖 ≤ 𝑥𝑖 (i.e., 𝑐𝑖 is in the ith subinterval) the Riemann sum is given by: 𝑛 𝑆𝑛 = ∑ 𝑓(𝑐𝑖 ) ∆𝑥 𝑖=1 The left sum is a Riemann sum in which every 𝑐𝑖 = 𝑥𝑖−1 (the left end of the subinterval): 𝑛 𝑆𝑛 = ∑ 𝑓(𝑥𝑖−1 ) ∆𝑥 𝑖=1 The right sum is a Riemann sum in which every 𝑐𝑖 = 𝑥𝑖 (the right end of the subinterval): 𝑛 𝑆𝑛 = ∑ 𝑓(𝑥𝑖 ) ∆𝑥 𝑖=1 The midpoint sum is a Riemann sum in which every 𝑐𝑖 = 𝑥𝑖 +𝑥𝑖−1 2 (the midpoint of the subinterval): 𝑛 𝑥𝑖 + 𝑥𝑖−1 𝑆𝑛 = ∑ 𝑓 ( ) ∆𝑥 2 𝑖=1 The Definite Integral If f(x) is continuous on [a,b] the definite integral of f(x) from a to b is the limit of any Riemann sum as n approaches infinity: 𝑛 𝑏 ∫ 𝑓(𝑥)𝑑𝑥 = lim 𝑆𝑛 = lim (∑ 𝑓(𝑐𝑖 ) ∆𝑥) 𝑎 𝑛→∞ 𝑛→∞ 𝑖=1 The definite integral represents the signed area between the graph of f(x) and the x-axis on the interval [a,b]. The area above the x-axis is positive and the area below the x-axis is negative. 4 From example 1, ∫2 2𝑥 𝑑𝑥 = 12 4 From example 2, ∫0 √4𝑥 − 𝑥 2 𝑑𝑥 = 2𝜋 Properties of Definite Integrals 𝑎 1. ∫ 𝑓(𝑥)𝑑𝑥 = 0 𝑎 𝑏 𝑎 2. ∫ 𝑓(𝑥)𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥 𝑎 𝑏 𝑏 𝑏 3. ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥 𝑎 𝑎 𝑏 𝑏 𝑏 4. ∫ [𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥) 𝑑𝑥 𝑎 𝑏 𝑎 𝑎 𝑐 𝑏 5. ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥 𝑎 𝑎 𝑐 Example 3. In the illustration to the right, area A is 44.8 square units, area B is 19.2 and area C is 83.2. Find 0 A) ∫−2(8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = −44.8 2 B) ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 19.2 4 C) ∫2 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 83.2 2 0 2 4 2 4 D) ∫−2(8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = ∫−2(8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 + ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = −44.8 + 19.2 = −25.6 E) ∫−2(8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = ∫−2(8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 + ∫2 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = −25.6 + 83.2 = 57.6 4 2 4 F) ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 + ∫2 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 19.2 + 83.2 = 102.4 0 2 G) ∫2 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = − ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = −19.2 −1 3 If ∫−2 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = −42.4, ∫1 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 63.2, 4 4 ∫1 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 100.8, and ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 102.4 then find H) 𝐴𝑟𝑒𝑎 𝐴 = 42.4 4 3 I) 𝐴𝑟𝑒𝑎 𝐵 = ∫1 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 − ∫1 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 100.8 – 63.2 = 37.6 1 4 4 J) ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = ∫0 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 − ∫1 (8𝑥 3 − 2𝑥 4 ) 𝑑𝑥 = 102.4 − 100.8 = 1.6 Example 4. In the illustration at the right, area A is 7.52, area B is 6.39, and area C is 0.68. Find 2 A) ∫0 5𝑒 −0.3𝑡 𝑑𝑡 = 7.52 6 B) ∫2 5𝑒 −0.3𝑡 𝑑𝑡 = 6.39 10 C) ∫8 5𝑒 −0.3𝑡 𝑑𝑡 = 0.68 6 2 6 D) ∫0 5𝑒 −0.3𝑡 𝑑𝑡 = ∫0 5𝑒 −0.3𝑡 𝑑𝑡 + ∫2 5𝑒 −0.3𝑡 𝑑𝑡 = 7.52 + 6.39 = 13.91 8 If ∫2 5𝑒 −0.3𝑡 𝑑𝑡 = 7.63 then find 8 8 6 E) ∫6 5𝑒 −0.3𝑡 𝑑𝑡 = ∫2 5𝑒 −0.3𝑡 𝑑𝑡 − ∫2 5𝑒 −0.3𝑡 𝑑𝑡 = 7.63 − 6.39 = 1.24 10 2 8 10 F) ∫0 5𝑒 −0.3𝑡 𝑑𝑡 = ∫0 5𝑒 −0.3𝑡 𝑑𝑡 + ∫2 5𝑒 −0.3𝑡 𝑑𝑡 + ∫8 5𝑒 −0.3𝑡 𝑑𝑡 = 7.52 + 7.63 + 0.68 = 15.83 Actually, 15.84 is a better approximation. Our value is a little off because of round-off errors when our areas were rounded down to two decimal places.