If sin(1/x^2+1) is an antiderivative for f(x) then.
1 answer:
The limit from 1 to 2 of the given antiderivative is; -0.19865
<h3>What is the Limit of the Integral?</h3>
We are given the antiderivative of f(x) as sin(1/(x² + 1)). Thus, to find the limit from 1 to 2, we will solve as;
⇒ (sin ¹/₅) - (sin ¹/₂)
⇒ 0.19866 - 0.47942
⇒ -0.19865
Complete Question is;
If sin(1/(x² + 1)) is an anti derivative for f(x), then what is the limit of f(x)dx from 1 to 2?
Read more about integral limits at; brainly.com/question/10268976
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The approximations T8 and M8 for the given integral are:
- T8 = 33.386321; and
- M8 = 33.50794
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: = Δx/2
= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]
= 0.902333
Now
T8 =
=
= 37 (0.902333)
T8 = 33.386321
It is to be noted that the midpoints rule is given as; = 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 =
=
= 37 (0.905620)
M8 = 33.50794
Learn more about integral at;
brainly.com/question/19053586
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