The average value of f(x) over [2, 6] is given by the definite integral,
[tex]\displaystyle f_{\rm ave[2,6]} = \frac1{6-2} \int_2^6 3\ln(2+x^3)\cos(x) \, dx[/tex]
and is approximately -1.67284.
The approximate average value of the function in the closed interval [2,6] is -1.628.
It is given that the f is the function given by: [tex]\rm f(x) = 3ln(2+x^2)cosx[/tex]
It is required to find the average value of f in the closed interval [2,6]
It is defined as the mathematical approach to calculating the smaller parts or components.
We have function f:
[tex]\rm f(x) = 3ln(2+x^2)cosx[/tex]
For the average value in [2,6]
We integrate the function with lower limit 2 and higher limit 6.
[tex]\rm \int_{2}^{6}f(x) =\int_{2}^{6}( 3ln(2+x^2)cosx)\\[/tex]
The average value of the above function:
[tex]\rm =\frac{1}{6-2} \int_{2}^{6}( 3ln(2+x^2)cosx)\\[/tex]
Further solving:
[tex]\rm =\frac{3}{4} \int_{2}^{6}( ln(2+x^2)cosx)\\[/tex]
Further solving and applying limits we get:
[tex]=\frac{3}{4}\times-2.2304[/tex]
= -1.628
Thus, the approximate value of the function in the closed interval [2,6]
is -1.628.
Learn more about the integration here:
https://brainly.com/question/14502499
