91Ó°ÊÓ

An antiderivative is a fundamental concept in calculus. It is a function that "undoes" differentiation, commonly referred to as the indefinite integral. When you differentiate a function and then take an antiderivative, you return to your original function, up to a constant. In essence, the antiderivative of a function is a new function whose derivative is the original function. Let's look at the antiderivative for the function given in the exercise:

• The function is given as \( f(x) = \frac{3}{64} x^{2} \).
• We seek an antiderivative \( F(x) \) such that when differentiated, \( F'(x) = f(x) \).
• For \( x^2 \), the antiderivative is \( \frac{x^3}{3} \), so by multiplying through by \( \frac{3}{64} \), the antiderivative becomes \( F(x) = \frac{x^3}{64} \).

The purpose of finding an antiderivative in this context is to prepare us for evaluating a definite integral. We'll use \( F(x) \) in the definite integration process over the interval \([0, 4]\).

A definite integral helps to find the area under a curve over a particular interval, specifically between two points on the x-axis. For the exercise, we use it to find the total area under \( f(x) \) from 0 to 4 and verify a property of probability density functions. The definite integral of \( f(x) = \frac{3}{64} x^2 \) from 0 to 4 is calculated as follows:

• Begin by finding the antiderivative, which we identified as \( \frac{x^3}{64} \).
• Evaluate this antiderivative at the bounds of our interval: \([0, 4]\).

With the formula for a definite integral, once you have \( F(x) = \frac{x^3}{64} \), you compute: \[ \left[ \frac{x^3}{64} \right]_0^4 = \frac{4^3}{64} - \frac{0^3}{64} \] This gives us \( \frac{64}{64} - 0 = 1 \), demonstrating that the total area under the curve within this interval is 1. This confirms that the function satisfies an essential property of probability density functions.

A probability density function (PDF) describes the likelihood of a random variable within a particular range. It has specific properties that must be satisfied: one of which is that the total area under the curve of a PDF over its defined interval must equal 1. This property ensures the function is properly normalized and can represent probabilities. In our exercise, this property is verified by:

• Integrating the function \( f(x) = \frac{3}{64} x^{2} \) over the interval \([0, 4]\).
• Computing the area using the definite integral. This requires the antiderivative, which converts the function describing the slope back into a function describing area under the curve.

Once we've carried out these steps and found the integral equals to 1, we affirm that \( f(x) \) is a valid probability density function over the specified interval. This validation is crucial as it allows \( f(x) \) to be used in statistical applications involving continuous data.