Question

For every one-dimensional set \(C\) for which the integral exists, let \(Q(C)=\) \(\int_{C} f(x) d x\), where \(f(x)=6 x(1-x), 0

Short answer

The integrals \(Q(C_{1})\), \(Q(C_{2})\), and \(Q(C_{3})\) need to be calculated using calculus techniques. Note however that \(Q(C_{2})\) is zero as it is an integral over a single point. Calculating these integrals will provide the required values of \(Q(C_{1})\), \(Q(C_{2})\), and \(Q(C_{3})\).

Step-by-step solution

Evaluate \(Q(C_{1})\)

The set \(C_1\) is \(\left\{x: \frac{1}{4}<x<\frac{3}{4} \right\}, also called an open interval. The function is defined within this interval, so we can calculate the integral: \(Q(C_{1})= \int_{1/4}^{3/4} 6x(1-x) dx\). This integral can be solved using standard calculus techniques.

Evaluate \(Q(C_{2})\)

The set \(C_2\) contains only one point \(x = 1/2\). The integral over an interval of zero width (a single point) is equal to zero, so we have \(Q(C_{2})=0\).

Evaluate \(Q(C_{3})\)

The set \(C_3\) is \(\{x: 0<x<10\}\). However, our function is only defined in the range 0 to 1, and is zero thereafter. Therefore, while this set extends from 0 to 10, in practical terms it is equivalent to the closed interval (0,1). We can calculate the integral as: \(Q(C_{3})= \int_{0}^{1} 6x(1-x) dx\). This integral can be calculated using standard calculus techniques.