Answer:
a) 72.5
b) 72.64
c) 72.66
Step-by-step explanation:
(a) 2 sub-intervals of equal length (width = 2/2 = 1) would have their mid point at x = 5.5 and x = 6.5.
[tex]f(5.5) = 5.5^2 = 30.25[/tex]
[tex]f(6.5) = 6.5^2 = 42.25[/tex]
So the sum over interval [5,7] is
30.25*1 + 42.35*1 = 72.5
(b) 5 sub-intervals of equal length (width = 2/5 = 0.4) would have their mid point at x = 5.2, 5.6, 6, 6.4, 6.8
[tex]f(5.2) = 5.2^2 = 27.04[/tex]
[tex]f(5.6) = 5.6^2 = 31.36[/tex]
[tex]f(6) = 6^2 = 36[/tex]
[tex]f(6.4) = 6.4^2 = 40.96[/tex]
[tex]f(6.8) = 6.8^2 = 46.24[/tex]
So the sum over interval [5,7] is
27.04*0.4 + 31.36*0.4 + 36*0.4 + 40.96*0.4 + 46.25*0.4 = 72.64
(c) 10 sub-intervals of equal length (width = 2/10 = 0.2) would have their mid point at x = 5.1, 5.3, 5.5, 5.7, 5.9, 6.1, 6.3, 6.5, 6.7, 6.9
[tex]f(5.1) = 5.1^2 = 26.01[/tex]
[tex]f(5.3) = 5.3^2 = 28.09[/tex]
[tex]f(5.5) = 5.5^2 = 30.25[/tex]
[tex]f(5.7) = 5.7^2 = 32.49[/tex]
[tex]f(5.9) = 5.9^2 = 34.81[/tex]
[tex]f(6.1) = 6.1^2 = 37.21[/tex]
[tex]f(6.3) = 6.3^2 = 39.69[/tex]
[tex]f(6.5) = 6.5^2 = 42.25[/tex]
[tex]f(6.7) = 6.7^2 = 44.89[/tex]
[tex]f(6.9) = 6.9^2 = 47.61[/tex]
So the sum over interval [5,7] is
0.2 * (26.01 + 28.09+30.25 + 32.49 + 34.81 + 37.21 + 39.69 + 42.25 + 44.89 + 47.61) = 72.66
