See attached, thanks!

The graph shows data trafc on the link from the United States to SWITCH, the Swiss education and research network, on February 10, 1998. D(t) is the data throughput, measured in megabits per second {Mbis}. Use Simpson's Rule to estimate the total amount of data transmitted on the link up to noon on that day. Because we want the units to be consistent and D(t) is measured in megabits per second, we convert the units for t from hours to seconds. If we let A(t) be the amount of data (in megabits) transmitted by time t, where t is measured in seconds, then A'(t) = D(t). So, by the Net Change Theorem, the total amount of data transmitted by noon (when t = 12 X (:102 = 43200) is 43200 A(43200) = f on) dt. 0 We estimate the values ofD(t) at hourly intervals from the graph and compile them in the table. Hhours) Hseconds} D(t) t(hours) Hseconds) D(t) 0 0 3.2 7 25,200 1.3 1 3,600 2.7 8 28,800 2.8 2 7,200 1 .9 9 32,400 5.7 3 10,800 1.7 10 36,000 7.1 4 14,400 1.3 1 1 39,600 7.7 5 18,000 1.0 12 43,200 7.9 6 21,600 1.1 Then we use Simpson's Rule with n = 12 and At = 3600 to estimate the integral. 43200 At [0 Am dt 3 ?[D(D) + 4D(3600) + 29(7200) + - .- + 4D(39600) + M43200\" a $32 + 4(2.7) + 2(1.9) + 4(1.7) + 2(1.3) + 4(1.0) + 2(1.1) + 4(1.3)_ =[:J Thus the total amount of data transmitted up to noon is about [3 megabits. (Round your answer to the nearest thousand.)