Respuesta :

Answer:

The average value of the function [tex]f(x) = 6 - x^{2}[/tex] over the interval [tex][-1,4][/tex] is [tex]\frac{5}{3}[/tex].

Step-by-step explanation:

The average value of a function over an interval is represented by this integral:

[tex]\bar y = \frac{1}{b-a}\cdot \int\limits^{b}_{a} {f(x)} \, dx[/tex]

Where:

[tex]a[/tex], [tex]b[/tex] - Lower and upper bounds of the interval, dimensionless.

[tex]f(x)[/tex] - Function, dimensionless.

If [tex]a = -1[/tex], [tex]b = 4[/tex] and [tex]f(x) = 6 - x^{2}[/tex], the average value of the function is:

[tex]\bar y = \frac{1}{4-(-1)}\int\limits^{4}_{-1} {6-x^{2}} \, dx[/tex]

[tex]\bar y = \frac{6}{5}\int\limits^{4}_{-1} \, dx - \frac{1}{5}\int\limits^{4}_{-1} {x^{2}} \, dx[/tex]

[tex]\bar y = \frac{6}{5}\cdot x |_{-1}^{4} - \frac{1}{15}\cdot x^{3}|_{-1}^{4}[/tex]

[tex]\bar y = \frac{6}{5}\cdot [4-(-1)]- \frac{1}{15}\cdot [4^{3}-(-1)^{3}][/tex]

[tex]\bar y = \frac{5}{3}[/tex]

The average value of the function [tex]f(x) = 6 - x^{2}[/tex] over the interval [tex][-1,4][/tex] is [tex]\frac{5}{3}[/tex].

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