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Alekssandra [29.7K]
2 years ago
15

GOLD IN THE SKY

SAT
1 answer:
tankabanditka [31]2 years ago
3 0
GOLD IN THE SKY - By Alan E. Nourse
We can see in the story that the setting is in a cockpit, in a lab, simulating a landing to Mars because of this paragraph; "Greg twisted around in the cockpit that had been built just big enough to hold him, and shifted his long legs against the brace-webbing, trying to get them comfortable." We can see that Greg Hunter is nervous, "His heart hammered in his throat, and his eyes ached fiercely, but he paid no attention. His finger crept to the air-speed indicator, then to the cut-off switch." The time period we can state is around the 1950s. It implied the first trip to Mars, "He heard the count-down monitor clicking in his ears, and his hands clenched into fists. How far from Mars would he be 10 minutes from now?" In conclusion, the time is believed to be sometime in the 1950s, and the setting is in a lab where Gregory Hunter is going on a trip to mars when something wrong has happened.
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37 cos(x2) dx 0 Do the following. (a) Find the approximations T8 and M8 for the given integral. (Round your answer to six decima
dusya [7]

The approximations T8 and M8 for the given integral are:

  • T8 = 33.386321; and
  • M8 = 33.50794

<h3>What is an integral?</h3>

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.

<h3>What is the explanation to above answer?</h3>

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

Learn more about integral at;
brainly.com/question/19053586
#SPJ1

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