The average value of the function g = ∫62(5f(x)+2)dx given that the average value of the function f on the interval 2≤x≤6 is 3 is equal to 1054 units.
A function is integrable if and only if the integral exists for a given interval. Given an interval [a, b], the average value of the function is calculated by the following expression:
[tex]\bar f = \frac{1}{b-a}\int\limits^b_a {f(x)} \, dx[/tex] (1)
Where:
In this question we must apply this definition and properties of the integral to determine the average value of a given expression:
[tex]\bar g = \frac{1}{6 - 2}\cdot \left[\int\limits^6_2 {62\cdot (5\cdot f(x) + 2)} \, dx \right][/tex]
[tex]\bar g = \frac{310}{6-2} \int\limits^6_2 {f(x)} \, dx + \frac{124}{6-2} \int\limits^6_2 \, dx[/tex]
[tex]\bar g = 310 \cdot 3 +124[/tex]
[tex]\bar g = 1054[/tex]
The average value of the function g = ∫62(5f(x)+2)dx given that the average value of the function f on the interval 2≤x≤6 is 3 is equal to 1054 units. [tex]\blacksquare[/tex]
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