Problem 17 - Calculus and its applications

Question

Find \(k\) such that each function is a probability density function over the given interval. Then write the probability density function. $$ f(x)=k, \quad[3,9] $$

Step-by-Step Solution

Answer

The PDF is \( f(x) = \frac{1}{6} \) for \( x \) in [3, 9].

Step 1: Understand the Conditions

We need to find the constant \( k \) such that the function \( f(x) = k \) is a probability density function (PDF) over the interval \([3, 9]\). For a function to be a PDF, the integral of \( f(x) \) over the interval must equal 1.

Step 2: Set Up the Integral

Since \( f(x) = k \) is constant, the integral from 3 to 9 becomes:\[\int_{3}^{9} k \, dx = 1\]This is because the total area under the PDF over the interval must be 1.

Step 3: Calculate the Integral

The integral of a constant \( k \) over an interval \([3, 9]\) is:\[\int_{3}^{9} k \, dx = k \times (9 - 3) = k \times 6\]Therefore, we need:\[k \times 6 = 1\]

Step 4: Solve for k

Solve the equation for \( k \):\[k \times 6 = 1 \k = \frac{1}{6}\]So, \( k = \frac{1}{6} \).

Step 5: Write the Probability Density Function

With the constant \( k \) found, the probability density function is:\[f(x) = \frac{1}{6}, \quad 3 \leq x \leq 9\]Thus, \( f(x) \) is a valid PDF over the interval \([3, 9]\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the areas under and between curves. In simpler terms, it helps us find totals where adding up tiny bits is easier than tackling the whole thing at once. This is especially handy when dealing with continuous functions.

In our context, when we talk about finding the probability density from a function across an interval, we use integration to "add up" the value of this function over the interval. Imagine a smooth layer of icing spread evenly on a cake, where you want to know how thick the icing is between two points on the cake's surface. Using integral calculus, you can determine that total thickness by adding up the icing's density across the interval.

• Important to note: Always remember, in integral calculations related to probability, the purpose is to find an accumulated quantity (often probability) over a range.
• Integration can require different techniques depending on the function, but for constant functions, it simplifies to straightforward multiplication.

Although integral calculus can seem complex, it breaks problems into manageable pieces by calculating the area under curves, ultimately making it valuable in probability contexts as well.

Definite Integral

The definite integral is a specific type of integration in mathematics where you compute the integral of a function over a particular interval, providing a precise numerical value. It's like calculating the total cost of groceries, where each item adds up to a total bill. The difference, however, is that you are dealing with functions instead of numbers.

In the scope of probability density functions (PDFs), the definite integral represents the total probability over a defined interval. For a function to qualify as a PDF, its definite integral over a specific interval must equal 1, indicating that there's a 100% chance the event or result falls within that range.

• The notation for definite integral involves specifying the function and the interval of interest, such as: \[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]where \(a\) and \(b\) are the interval limits, and \(F(x)\) is the antiderivative of \(f(x)\).
• Definite integrals are widely used in various fields, particularly in physics and engineering, to calculate things like work done by forces or the flow of liquids.

In summary, the definite integral gives a concrete answer about the size or magnitude of a phenomenon over a range, which, in the case of PDFs, translates to the certainty of outcomes within an interval.

Constant Function as PDF

When dealing with probability density functions (PDFs), a constant function represents a scenario where the probability is uniformly distributed across a defined interval. Essentially, this means the chance of any outcome happening within the interval is equally likely.

In our exercise example, the function is defined as \( f(x) = k \), where \(k\) is a constant, and the interval is \([3,9]\). The challenge is to find the appropriate \(k\) that makes the integral of this constant function over the interval equal to 1, thereby satisfying the requirement for it to be a valid PDF.

• A constant function has a 'flat line' graph, representing an even chance of any event happening over the interval.
• The process involves integrating this function over the specified interval, multiplying the constant by the length of the interval.
• In the solution, \(k = \frac{1}{6}\) was determined because it balances the function to meet the total probability of 1 across the interval.

Overall, constant functions simplify the process of creating PDFs, as they provide a straightforward way to represent equal probability across outcomes.