The average value of a function g on 0 F$\le$x $\le$3 is a constant $\bar{g}$ given by $\bar{g}=\frac{1}{3-0} \int_{0}^{3} g(x) d x$ .Which of the following must be true? A) $\int_{0}^{3} \bar{g} g(x)(\bar{g})^{2}$ C) $\int_{0}^{3} \bar{g} g(x)=(\bar{g})^{2}$ D)None of the above

Question

The average value of a function g on 0 F$\le$x $\le$3 is a constant $\bar{g}$ given by $\bar{g}=\frac{1}{3-0} \int_{0}^{3} g(x) d x$ .Which of the following must be true?
A) $\int_{0}^{3} \bar{g} g(x)<(\bar{g})^{2}$
B) $\int_{0}^{3} \bar{g} g(x)>(\bar{g})^{2}$
C) $\int_{0}^{3} \bar{g} g(x)=(\bar{g})^{2}$
D)None of the above

Quizplus · Accepted Answer

The average value of a function is equal to the integral of the function divided by the length of the interval over which the function is defined. In this case, the average value of g on [0, 3] is $\bar{g}=\frac{1}{3-0} \int_{0}^{3} g(x) d x$. The integral of the product of two functions is equal to the product of the integrals of the two functions. Therefore, $\int_{0}^{3} \bar{g} g(x) = \bar{g} \int_{0}^{3} g(x) d x = \bar{g} \cdot 3 \bar{g} = (\bar{g})^{2}$.

The Average Value of a Function G on 0 F ≤\le≤

The Average Value of a Function G on 0 F $\le$