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Chapter 13 – 4a The Definite Integral Approximating Areas Using Left and Right Sums y = 2x 12 Example 1. Find the approximate area between the graph of y = 2x and the xaxis from x = 2 to x = 4. The actual area is 12. 10 8 6 4 2 Using the sum of the areas of rectangles whose heights are the values of y for the left x-coordinate: 1 1 1 0 0 1 2 3 4 5 1 π΄πππ ≈ 4 β 2 + 5 β 2 + 6 β 2 + 7 β 2 y = 2x 12 1 π΄πππ ≈ (4 + 5 + 6 + 7) β 2 = 11 10 8 6 4 2 0 0 1 2 3 4 5 Using the sum of the areas of rectangles whose heights are the values of y for the right x-coordinate: 1 1 1 1 π΄πππ ≈ 5 β 2 + 6 β 2 + 7 β 2 + 8 β 2 y = 2x 12 1 π΄πππ ≈ (5 + 6 + 7 + 8) β 2 = 13 10 8 6 4 2 0 0 1 2 3 4 5 To improve the accuracy of the approximation, increase the number of rectangles. Suppose we use 8 rectangles instead of 4: y = 2x y = 2x 12 12 10 10 8 8 6 6 4 4 2 2 0 0 0 1 2 3 4 5 0 1 2 3 4 5 1 4 Left Sum: π΄πππ ≈ (4 + 4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5) β = 11.5 1 Right Sum: π΄πππ ≈ (4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5 + 8) β 4 = 12.5 Example 2. Find the approximate area between the graph of π¦ = √−π₯ 2 + 2π₯ + 3 and the x-axis from x = 0 to x = 2. The region is a semi-circle 1 2 y = sqrt(4x - x2) 2 and the actual area is ππ 2 = 2π ≈ 6.283. 1 y = sqrt(4x - x2) 0 0 1 2 2 1 0 0 1 2 3 4 Using the left sum: π΄πππ ≈ 0 β 1 + 1.732051 β 1 + 2 β 1 + 1.732051 β 1 =(0 + 1.732051 + 2 + 1.732051) β 1 = 5.464 Using the right sum: y = sqrt(4x - x2) 2 1 0 0 1 2 3 4 π΄πππ ≈ 1.732051 β 1 + 2 β 1 + 1.732051 β 1 + 0 β= (1.732051 + 2 + 1.732051 + 0) β 1 = 5.464 y = sqrt(4x - x2) y = sqrt(4x - x2) 2 2 1 1 0 0 0 1 2 3 4 0 1 2 3 4 πΏπππ‘ ππ’π ≈ (0 + 1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32) β 1 = 5.991 2 π ππβπ‘ ππ’π ≈ (1.32 + 1.73 + 1.94 + 2.00 + 1.94 + 1.73 + 1.32 + 0) β 1 = 5.991 2 3 4 Riemann Sum The left sum and the right sum are 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 widthβπ₯ = π−π . π These subintervals have endpoints π₯0 , π₯1 , π₯2 , π₯3 , … , π₯π where π₯0 = π, π₯π = π, and π₯π − π₯π−1 = βπ₯ for any value of i from 1 to n. (This implies that π₯π = π + πβπ₯ for any i from 1 to n.) If π₯π−1 ≤ ππ ≤ π₯π (i.e., ππ is in the ith subinterval) the Riemann sum is given by: π ππ = ∑ π(ππ ) βπ₯ π=1 The left sum is a Riemann sum in which every ππ = π₯π−1 (the left end of the subinterval): π ππ = ∑ π(π₯π−1 ) βπ₯ π=1 The right sum is a Riemann sum in which every ππ = π₯π (the right end of the subinterval): π ππ = ∑ π(π₯π ) βπ₯ π=1 The midpoint sum is a Riemann sum in which every ππ = π₯π −π₯π−1 2 (the midpoint of the subinterval): π π₯π − π₯π−1 ππ = ∑ π ( ) βπ₯ 2 π=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: π π ∫ π(π₯)ππ₯ = lim ππ = lim (∑ π(ππ ) βπ₯) π π→∞ π→∞ π=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.