Respuesta :

calculista
we have 
y=cos x/(x²+x+2)  on the closed interval [-1, 3]

we know that
The average value of f(x) on the interval [a, b] is given by: 
F(avg) = 1/(b - a) ∫ f(x) dx (from x=a to b). 
(b-a)=(3+1)------> 4
= 1/4 ∫ cos(x)/(x² + x + 2) dx (from x=-1 to 3). 

Note that [cos(x)/(x² + x + 2)] does not have an elementary anti-derivative.
 By approximating techniques: 
1/4 ∫ cos(x)/(x^2 + x + 2) dx (from x=-1 to 3) ≈ 0.182951

the answer is
the average value of y = cos(x)/(x² + x + 2) on [-1, 3] is approximately 0.182951

MrRoyal

The average value of the function on the closed interval is 0.1428

The function is given as:

  • [tex]y = \frac{\cos(x)}{x^2 + x + 2}[/tex]
  • The interval is given as: [-1. 3]

Calculate the value of the function at points x = -1 and x = 3

[tex]y_{-1} = \frac{\cos(-1)}{(-1)^2 + (-1) + 2}[/tex]

[tex]y_{-1}= \frac{0.9998}{2}[/tex]

[tex]y_{-1}= 0.4999[/tex]

[tex]y_3 = \frac{\cos(3)}{3^2 + 3 + 2}[/tex]

[tex]y_3 = \frac{0.9986}{14}[/tex]

[tex]y_3 = 0.0713[/tex]

The average value of the function is:

[tex]Average = \frac{y_3 + y_{-1}}{3--1}[/tex]

[tex]Average = \frac{y_3 + y_{-1}}{3+1}[/tex]

This gives

[tex]Average = \frac{0.4999 + 0.0713}{4}[/tex]

[tex]Average = \frac{0.5712}{4}[/tex]

[tex]Average = 0.1428[/tex]

Hence, the average value of the function on the closed interval is 0.1428

Read more about average values at:

https://brainly.com/question/13136492

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