Which units are most appropriate for measuring the distance from the Earth to the sun?
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The value of the given statement according to the mid point theorem is 5.
According to the statement
we have given that the equation and we have to integrate it with the help of the mid point theorem.
So, For this purpose, we know that the
The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
So, The given equation is
where n = 4
And
The mid point formula is a with limits a to b is
f(x) dx ≈ Δx (f (x₀ + x₁)/2) + (f (x₁ + x₂)/2) + ... (f (xₓ₋₂ + xₓ₋₁)/2) + (f (xₓ₋₁ + x)/2))
Then
Where Δx = (b-a)/n
Recall that
a= 0
b= 64 and
n=4, therefore,
Δx = (64-0)/4 = 16
The next step requires that the interval [0,64] be divided into 4 sub-intervals with length = Δx =16
Hence, we've got, 0, 16, 32, 48, 64.
From this point, we calibrate the respective functions as follows:
f (x₀ + x₁)/2) = f ((0+16)/2) = f(8) = Sin (8) = 0.98935824662
(f (x₁ + x₂)/2) = f((16+32)/2) = f(24) =Sin (24) = -0.905578362
(f (x₂ + x₃)/2) = f((32+48)/2) = f(40) = Sin (40) = 0.74511316047
(f (x₃ + x₄)/2) = f((48+64)/2) = f(56) = Sin (56) = -0.52155100208
At this point, we sum up the above values derive the product of the total and Δx = 16
sin (x) dx = 16(0.98935824662 - 0.905578362 + 0.74511316047 - 0.52155100208)
= 16 (0.30734204301)
= 4.91747268816
So, The value of the given statement according to the mid point theorem is 5.
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