The average value of a function g on 0 $\le$x $\le$2 is a constant $\bar{g}$ given by $\bar{g}=\frac{1}{2-0} \int_{0}^{2} g(d) d x$ .Also, $\int_{0}^{2}(g(x)-\bar{g})^{2} d x \geq 0$ , since $(g(x)-\bar{g})^{2}$ is a square.Which of the following must be true?
A) $\left(\int_{0}^{2} g(x) d x\right)^{2} \leq \int_{0}^{2}(g(x))^{2} d x$
B) $\left(\int_{0}^{2} g(x) d x\right)^{2} \geq \int_{0}^{2}(g(x))^{2} d x$
C) $\left(\int_{0}^{2} g(x) d x\right)^{2}=\int_{0}^{2}(g(x))^{2} d x$
D)None of these

Question

Quizplus · Accepted Answer

The inequality $\left(\int_{0}^{2} g(x) d x\right)^{2} \leq \int_{0}^{2}(g(x))^{2} d x$ is a consequence of the Cauchy-Schwarz inequality for integrals, which states that the square of the integral of a product of functions is less than or equal to the product of the integrals of the squares of the functions. This principle implies that the square of the integral of a function (which can be seen as a dot product in function space) is less than or equal to the integral of the square of the function, reflecting the geometric interpretation of the Cauchy-Schwarz inequality.

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

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