Answer:
The average value of [tex]f[/tex] over the interval [tex][-2,5][/tex] is [tex]\frac{10}{7}[/tex].
Step-by-step explanation:
Let suppose that function [tex]f[/tex] is continuous and integrable in the given intervals, by integral definition of average we have that:
[tex]\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6[/tex] (1)
[tex]\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20[/tex] (2)
By Fundamental Theorems of Calculus we expand both expressions:
[tex]\frac{F(3)-F(-2)}{3-(-2)} = -6[/tex]
[tex]F(3) - F(-2) = -30[/tex] (1b)
[tex]\frac{F(5)-F(3)}{5-3} = 20[/tex]
[tex]F(5) - F(3) = 40[/tex] (2b)
We obtain the average value of [tex]f[/tex] over the interval [tex][-2, 5][/tex] by algebraic handling:
[tex]F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)[/tex]
[tex]F(5) - F(-2) = 10[/tex]
[tex]\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}[/tex]
[tex]\bar f = \frac{10}{7}[/tex]
The average value of [tex]f[/tex] over the interval [tex][-2,5][/tex] is [tex]\frac{10}{7}[/tex].
