Integration - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {9}Approximating the area under the graph with 10 rectanglesInterval Width of rectangle Height of rectangle Area of rectangle[0, 1 2 ] 12f (0) = 0 + 1 = 1[ 1 2 ,1] 12f ( 1 2 ) = 1 4 + 1 = 5 4[1, 3 2 ] 12f (1) = 1 + 1 = 2 1[ 3 2 ,2] 12f ( 3 2 ) = 9 4 + 1 = 13 4[2, 5 2 ] 12f (2) = 4 + 1 = 5[ 5 2 ,3] 12f ( 5 2 ) = 25 4 + 1 = 29 4[3, 7 2 ] 12f (3) = 9 + 1 = 10 5[ 7 2 ,4] 12f ( 7 2 ) = 49 4 + 1 = 53 4[4, 9 2 ] 12f (4) = 16 + 1 = 17[ 9 2 ,5] 12f ( 9 2 ) = 81 4 + 1 = 85 4125813852298538172858Total area3258 = 40.625Splitting [0,5] into more subintervals of smaller width, we can perform the same calculationand estimate the area. With more rectangles, the calculations become more tedious,but some results are given in the following table.Approximating the area under the graph with more rectanglesNumber of intervals Width of each interval Total area of rectangles5 1 3510 0.5 40.62550 0.1 45.425100 0.05 46.043751 000 0.005 46.604187510 000 0.0005 46.66041875100 000 0.00005 46.666041668751 000 000 0.000005 46.6666041666875With more and more thinner and thinner rectangles, the areas appear to be convergingtowards a limit. It turns out, as we will see, that the limit is 46 2 3 .The area under a curve is defined to be this limit. Although we have an intuitive notionof what area is, for a mathematically rigorous definition we need to use integration.