91Ó°ÊÓ

When working with probability density functions (pdfs), it's crucial to ensure the function in question is non-negative over its entire domain. A non-negative function is one that never dips below the horizontal axis; it always takes on values greater than or equal to zero. For a function to represent a pdf, it must be non-negative because the probability of any event cannot be negative.

In our example, the function given is \( f(x)=4x^3 \). To confirm that this function is non-negative on the interval [0,1], we recognize that any real number raised to the third power retains its sign. Since we're only dealing with the interval from zero to one, all our x-values are already non-negative. Multiplying these by 4 will still keep them non-negative, thus fulfilling one of the conditions to be a valid pdf.

The integral of a function can be thought of as the total accumulated value of the function over a certain interval. For pdfs, calculating the integral of the function over its entire domain tells us the total probability, which must be 1, since the probability of all possible outcomes combined is certain.

In practical terms, the integral represents the area under the curve of the function for a specified interval. In the given exercise, we need to determine if the total area under the curve of \( f(x)=4x^3 \) from 0 to 1 is equal to 1. If it is, this signifies that the function can represent all possible outcomes in a probabilistic sense, which is a necessity for any pdf.

To calculate integrals effectively, we often rely on integration rules, such as the power rule for integration. This rule simplifies the integration process for polynomials. The rule states that the integral of \( x^n \) with respect to x is \( \frac{x^{n+1}}{n+1} \), plus a constant of integration when an indefinite integral is performed.

In our case, when applying the power rule to integrate \( 4x^3 \) from 0 to 1, we increase the exponent by one and divide by the new exponent. Thus, the integral becomes \( \frac{4x^{3+1}}{3+1} \) which simplifies to \( x^4 \). After integrating, we evaluate this expression from 0 to 1, confirming that the area under the curve is indeed 1, which is the confirmation needed for \( f(x)=4x^{3} \) to be a valid pdf over the interval [0,1].