Question

37 cos(x2) dx 0 Do the following. (a) Find the approximations T8 and M8 for the given integral. (Round your answer to six decimal places.)

Answer 1

The approximations T8 and M8 for the given integral are:

• T8 = 33.386321; and
• M8 = 33.50794

What is an integral?

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.

What is the explanation to above answer?

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:

$\int\limits^1_0 cos(x)^{2} } \, dx$ = Δx/2 $[f(xo) + 2f(x1) 2f(x2) + ....+ 2f(x7) + f(x8)]$

= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]

= 0.902333

Now

T8 = $\int\limits^1_0 {37Cos(x)^{2} } \, dx$

= $37\int\limits^1_0 {(0.902333)} } \, dx$

= 37 (0.902333)

T8 = 33.386321

It is to be noted that the midpoints rule is given as;

$\int\limits^1_0 {Cos(x)^{2} } \, dx$  = 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 = $\int\limits^1_0 {37 Cos(x)^{2} } \, dx$

= $37\int\limits^1_0 {Cos(x)^{2} } \, dx$

= 37 (0.905620)

M8 = 33.50794

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