1 1 of 19 (1 complete) a) Approximate the area under graph (a) of t(x)-1/x over
ID: 2890108 • Letter: 1
Question
1 1 of 19 (1 complete) a) Approximate the area under graph (a) of t(x)-1/x over the interval [2,10] by computing the area of each rectangle to four decimal places and then adding b) Approximate the area under graph (b) of 10) 1/x over the inferval 12, 10] by computing the area of each rectangle to four decimal places and then adding The area urder graph (a) is approximately (Round to four decimal places as needed.) 0.125 The area under graph (b) is approximately (Round to four decimal places as needed.) 10 0.1 3 4 56 7 8 9 10Explanation / Answer
a)
Height of first rectangle=f(2)=0.25 => Area=0.25*2=0.5
Height of second rectangle=f(4)=0.125/2 => Area=0.125/2*2=0.125
Height of third rectangle=f(6)=0.25/9 => Area=0.25/9*2=0.5/9=0.0555
Height of fourth rectangle=f(8)=0.25/16 => Area=0.25/16*2=0.5/16=0.03125
Total area=0.5+0.125+0.0555+0.03125=0.71175
b)
Height of first rectangle=f(2)=0.25 => Area=0.25*1=0.25
Height of second rectangle=f(3)=0.25/4 => Area=0.25/4*1=0.0625
Height of third rectangle=f(4)=0.25/9 => Area=0.25/9*1=0.25/9=0.0277
Height of fourth rectangle=f(5)=0.25/16 => Area=0.25/16*1=0.125/8=0.015625
Height of fourth rectangle=f(6)=0.25/25 => Area=0.25/25*1=0.01
Height of fourth rectangle=f(7)=0.25/36 => Area=0.25/36*1=0.00694
Height of fourth rectangle=f(8)=0.25/49 => Area=0.25/49*1=0.0051
Height of fourth rectangle=f(6)=0.25/81 => Area=0.25/81*1=0.00308
Total area=0.3810
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