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For a function to be considered a probability density function (PDF), it needs to meet the criterion of non-negativity. This means that the function must have non-negative values for all possible input values of the random variable. In simple terms, a PDF cannot be negative because probabilities cannot be negative. Consider the function given in the exercise, where \( p(x) = xe^{-x^2} \) for \( x > 0 \) and \( p(x) = 0 \) for \( x \leq 0 \).

When \( x > 0 \), both \( x \) and \( e^{-x^2} \) are positive. Multiplying these positive values ensures that \( p(x) \) remains non-negative. Moreover, for \( x \leq 0 \), \( p(x) \) is defined as 0, which is non-negative. Therefore, in all cases, \( p(x) \geq 0 \), fulfilling the non-negativity requirement for a PDF.

One essential criterion for a probability density function is that its integral over the entire possible range should equal 1. This ensures that the total probability sums up to 100%, as probabilities are always fractions of a whole. Sometimes, we deal with integrals that have infinite limits, known as improper integrals.

In the provided exercise, the function \( p(x) \) is defined only for \( x > 0 \). We need to compute the integral of \( p(x) = xe^{-x^2} \) over this range, denoted as \( \int_{0}^{\infty} xe^{-x^2} \, dx \).

Solving this involves evaluating an improper integral, as the upper limit extends to infinity. After calculating, we find the resulting integral equals \( \frac{1}{2} \). This result shows the total integrated probability does not equal 1, but instead 0.5, indicating \( p(x) \) does not satisfy the integral property of a probability density function.

Integration by substitution is a handy mathematical technique often used to simplify complicated integrals. In essence, it allows us to swap out variables to make the integral easier to manage. It's like solving a puzzle by rearranging the pieces to reveal the picture.

In this case, we have the integral \( \int_{0}^{\infty} xe^{-x^2} \, dx \). At first glance, this is a bit complex to solve directly. To ease the process, we use substitution. Let us set a substitution where \( u = x^2 \), leading to \( du = 2x \, dx \), implying \( x \, dx = \frac{1}{2} \, du \).

Thus, the integral transforms to \( \frac{1}{2} \int_{0}^{\infty} e^{-u} \, du \). This simplified integral is much easier to evaluate. The integral of \( e^{-u} \) from 0 to infinity is a common result equalling 1. Multiplying by the \( \frac{1}{2} \) factor from our substitution yields \( \frac{1}{2} \).

This process elegantly shows how substitution can transform an intimidating integral into a manageable form. However, in the context of the original problem, the final integrated value of \( \frac{1}{2} \) indicates that \( p(x) \) isn't a true probability density function.