91Ó°ÊÓ

A Probability Density Function (PDF) must always be nonnegative for all values within its interval. This means that a PDF, like the function given, must satisfy the condition \(f(x) \geq 0\) for any \(x\) in the interval it covers. In the exercise, the interval given is \([0, 1]\).

By sampling points within the interval, such as \(x = 0\), \(x = 0.5\), and \(x = 1\), we can demonstrate nonnegativity. When each value of \(x\) is substituted back into the function \(f(x) = x + 2x^{3}\), it results in a nonnegative number:

• \(f(0) = 0 + 2 \cdot 0^{3} = 0\)
• \(f(0.5) = 0.5 + 2 \cdot (0.5)^{3} = 0.5 + 0.25 = 0.75\)
• \(f(1) = 1 + 2 \cdot 1^{3} = 1 + 2 = 3\)

The function remains nonnegative across the entire interval (including when \(x\) is not a simple fraction). This confirms that the nonnegativity condition for a PDF is met.

Understanding the definite integral is essential for confirming that a function can serve as a probability density function over an interval. The definite integral calculates the 'area under the curve' of a function for a specific range of \(x\), signified as \([a, b]\). In this exercise, we act within the interval \([0, 1]\).

To verify that a function is a PDF, we compute \(\int_{0}^{1} f(x) \, dx\). For a genuine PDF, this integral must equal 1, representing a total probability of 1 over the interval.

For the function \(f(x) = x + 2x^{3}\), the definite integral is given by:
\[\int_{0}^{1} (x + 2x^3) \, dx\]
The value of this definite integral reflects if the total probability of all possible outcomes on the interval sums to 1. Solving this integral as demonstrated in the following section will show that the function satisfies this critical requirement for a PDF.

The power rule for integrals is a basic but powerful tool for computing integrals, essential when working with polynomial functions. The rule states that the integral of \(x^n\) with respect to \(x\) is \(\frac{1}{n+1}x^{n+1}\) plus a constant of integration (which is omitted when calculating definite integrals over a finite interval).

In our exercise, the power rule is applied to find the definite integral of \(f(x) = x + 2x^3\) from \(0\) to \(1\). Applying the power rule, we find:

• \(\int x \, dx = \frac{1}{2}x^2\)
• \(\int 2x^3 \, dx = \frac{2}{4}x^4 = \frac{1}{2}x^4\)

Combining both parts, the definite integral simplifies to:
\[ \left.\left(\frac{1}{2}x^{2} + \frac{1}{2}x^{4}\right)\right|_{0}^{1} = (\frac{1}{2} \cdot 1^{2} + \frac{1}{2} \cdot 1^{4}) - (\frac{1}{2} \cdot 0^{2} + \frac{1}{2} \cdot 0^{4})\]

After calculating, we have:
\[ \frac{1}{2} + \frac{1}{2} = 1\]
The value \(1\) confirms that the integral satisfies the condition necessary for the function to be a valid PDF. This rule simplifies what might seem complex into a straightforward calculation, reinforcing why the power rule is a fundamental aspect of integration.