Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the value of \(k\) that makes \(f(x)=k x(5-x)\), \(0 \leq x \leq 5\), a valid PDF. Hint: The PDF must integrate to 1 .

Step-by-Step Solution

Answer

The value of \( k \) is \( \frac{6}{125} \).

Step 1: Understanding the Problem

To make a function a valid probability density function (PDF), it must integrate to 1 over a given interval. In this case, we need to set the integral of \( f(x) = kx(5-x) \) from 0 to 5 equal to 1 and solve for \( k \).

Step 2: Set Up the Integral

Set up the integral of the function \( f(x) = kx(5-x) \) over the interval from 0 to 5. This gives us the integral equation:\[\int_{0}^{5} kx(5-x) \, dx = 1\]

Step 3: Calculate the Integral

Expand \( x(5-x) \) to get \( 5x - x^2 \). Integrate this function with respect to \( x \):\[\int (5x - x^2) \, dx = \left[ \frac{5x^2}{2} - \frac{x^3}{3} \right]\]Evaluate the antiderivative from 0 to 5.

Step 4: Evaluate the Integral

Plug the limits into the antiderivative:\[\left[ \frac{5(5)^2}{2} - \frac{(5)^3}{3} \right] - \left[ \frac{5(0)^2}{2} - \frac{(0)^3}{3} \right] = \frac{125}{2} - \frac{125}{3}\]Calculate the result of this evaluation.

Step 5: Simplify the Integral Result

Combine the terms:\[\frac{125}{2} - \frac{125}{3} = \frac{375}{6} - \frac{250}{6} = \frac{125}{6}\]This is the value of the integral without \( k \). To make the function a valid PDF, we need this value times \( k \) to equal 1.

Step 6: Solve for \( k \)

Set the equation \( k \times \frac{125}{6} = 1 \) to solve for \( k \):\[k = \frac{6}{125}\]

Step 7: Conclusion

Thus, the value of \( k \) that makes the function a valid PDF is \( \frac{6}{125} \).

Key Concepts

Integral Calculus

Integral calculus is all about understanding how to calculate areas under curves. Imagine the curve as a stretched rubber sheet. We want to find the 'weight' underneath it, which tells us how likely certain outcomes are. In the given problem, this curve is expressed as the function

• \( f(x) = kx(5-x) \) over the interval \( 0 \leq x \leq 5 \).

By integrating this function, we find its total 'weight' or area. And for a probability density function (PDF), this must always add up to 1. So, integral calculus is crucial when proving whether a function like this is legit as a PDF. Let's walk through it: we set up the integral\[ \int_{0}^{5} kx(5-x) \, dx = 1 \] This expression ensures that the complete area (or probability) beneath the function from 0 to 5 sums to 1. That's how we know if a function does its job as a PDF.

Solving Equations

Solving equations in this context is about finding the correct multiplier \( k \) to satisfy the condition that makes the given function a valid probability density function. After setting up our integral and calculating it, we reached an expression

• \( k \times \frac{125}{6} = 1 \)

This equation is like a balance scale where the left side needs to equal the right side to keep things in balance, or in this case, to equal 1. By isolating \( k \) on one side of the equation, we find it using simple algebraic manipulation:\[ k = \frac{6}{125} \]This operation is a straightforward approach to finding the variable that binds the PDF together correctly over its domain.

Antiderivative

The antiderivative plays a pivotal role in solving this problem. It's the reverse process of differentiation, allowing us to determine the accumulated change across an interval. For the function

• \( 5x - x^2 \),

we find its antiderivative:\[ \frac{5x^2}{2} - \frac{x^3}{3} \] Evaluating this from the limits 0 to 5 gives the specific area under the curve. The concrete values used in our antiderivative help us calculate the definitive result

• \( \frac{125}{6} \)

for the integral without \( k \). Using antiderivatives is like uncovering the background story behind the instantaneous rates of changes represented by derivatives, piecing everything together across a stretch.

PDF Properties

PDF, or probability density function, must satisfy a few key properties to be valid. First, it must be non-negative for all values in its domain; this makes sense since negative probabilities don't have a physical meaning. More importantly, the total area under the curve of a PDF, when integrated over its entire range, must be 1. This entire area represents the total probability, encompassing all possible outcomes. In our exercise, we confirmed this by setting up the requirement:

• \( \int_{0}^{5} kx(5-x) \, dx = 1 \)

By ensuring the integral equals 1, we've confirmed the given function can operate as a bonafide PDF with the determined value of \( k = \frac{6}{125} \). These properties ensure that no probability is missing or extraneous, keeping the function aligned within the framework of probability.