91Ó°ÊÓ

An integral, in calculus, represents the area under a curve defined by a function, over a certain interval. It is a fundamental concept for understanding how quantities accumulate over a specified range. For instance, in probability theory, the integral of a probability density function (pdf) across its entire range should equal 1, symbolizing the certainty that a value within the range will occur.

To calculate the integral of a function, you often perform a process known as integration, which is, in some sense, the reverse operation of differentiation. There are two main types of integrals: indefinite and definite. An indefinite integral gives a family of functions and includes an arbitrary constant, representing the antiderivative of the original function. A definite integral, which we are concerned with in this exercise, has limits of integration and results in a numerical value. It determines the total accumulation between the specified bounds.

In the solution to our textbook problem, the definite integral \[\int_0^{\pi} f(x) \, dx\] is computed to confirm that the given function \(f(x)=\frac{1}{2} \sin x\) can serve as a pdf over the interval \[0, \pi\]. The evaluation of this integral provides the total 'area' under the function from \(x = 0\) to \(x = \pi\), which is a crucial step in probabalistic models.

A non-negative function is one that does not take on any negative values in its domain. In the context of probability, the notion of a non-negative function is important because probabilities cannot be negative - an event either happens or does not, with the probability reflecting a measure of certainty rather than a measure with direction.

For a function to be considered a probability density function, it must be non-negative over its entire domain. This means that for any value of \(x\), the function \(f(x)\) should be greater than or equal to zero. In the given textbook problem, the function \(f(x) = \frac{1}{2} \sin x\) meets this criterion on the interval \[0, \pi\] because the sine function oscillates between -1 and 1, hence, multiplying it by \(\frac{1}{2}\) ensures that it remains in the range \[0, 0.5\] for the specified interval. Establishing that \(f(x)\) satisfies the non-negativity requirement is essential before it can be recognized as a valid pdf.

Interval evaluation is a process used in calculus to find the value of a definite integral over a particular interval. It often involves determining the antiderivative of the function at the upper and lower bounds of the interval and finding the difference between these values. This difference represents the net 'area' or the 'accumulated quantity' under the curve between the two points.

In the solution provided, interval evaluation is carried out to prove that the integral of the function \(f(x) = \frac{1}{2} \sin x\) over the interval \[0, \pi\] is equal to 1. The antiderivative \[ - \frac{1}{2} \cos x \] is evaluated at the interval's end points creating \( - \frac{1}{2} \cos(\pi)\) and \( - \frac{1}{2} \cos(0)\). By calculating the difference, \(\frac{1}{2} - ( - \frac{1}{2})\), we find the result of the integral to be 1. This final step confirms that the area under the curve, or the total probability, is exactly 1, which is a critical property for any probability density function. The careful evaluation of intervals is necessary for a wide range of applications, not only in probability but also in physics, engineering, and economics.