The approximations T8 and M8 for the given integral are:
An Integral is a variable of which a given function is the derivative, i.e. it gives that function when differentiated and may express the area under the curve of the function's graph.
Given:
F(x) = 37 cox (x²)
Internal = [0,1] n = 8 in Δ x = 1/8
The sub intervals are:
[0, 1/8], [1/8, 2/8], [2/8, 3/8], [ 3/8, 4/8], [ 4/8, 5/8], [ 5/8, 6/8], [6/8, 7/8], [ 7/8, 1]
The mid points are given as:
1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16
and X₀ = 0, X₁ = 1/8, X₂ = 2/8
Using the Trapezium Rule which states that:
[tex]\int\limits^1_0 cos(x)^{2} } \, dx[/tex] = Δx/2 [tex][f(xo) + 2f(x1) 2f(x2) + ....+ 2f(x7) + f(x8)][/tex]
= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]
= 0.902333
Now
T8 = [tex]\int\limits^1_0 {37Cos(x)^{2} } \, dx[/tex]
= [tex]37\int\limits^1_0 {(0.902333)} } \, dx[/tex]
= 37 (0.902333)
T8 = 33.386321
It is to be noted that the midpoints rule is given as;
[tex]\int\limits^1_0 {Cos(x)^{2} } \, dx[/tex] = Δx [f(1/16) + (3/16) + .... + f(15/16)]
= 1/8[f(1/16) + f (3/16) + f(5/16) + f(7/16) + f(9/16) + f(11/16) + f(13/16) + f(15/16)]
= 0.905620
From the above,
M8 = [tex]\int\limits^1_0 {37 Cos(x)^{2} } \, dx[/tex]
= [tex]37\int\limits^1_0 {Cos(x)^{2} } \, dx[/tex]
= 37 (0.905620)
M8 = 33.50794
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