Find the value of the constant c so that the given function is a probability density function for a random variable X over the specified interval.
f(x) = c * x * √(25 - x²) over [0, 5]

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Recall that for a function \(f(x)\) to be a probability density function (pdf) over an interval \([a, b]\), the total area under the curve must equal 1. This means we need to solve the equation \(\int_a^b f(x) \, dx = 1\).

Set up the integral for the given function \(f(x) = c \cdot x \cdot \sqrt{25 - x^2}\) over the interval \([0, 5]\): \(\int_0^5 c \cdot x \cdot \sqrt{25 - x^2} \, dx = 1\).

Since \(c\) is a constant, factor it out of the integral: \(c \cdot \int_0^5 x \cdot \sqrt{25 - x^2} \, dx = 1\).

Focus on evaluating the integral \(\int_0^5 x \cdot \sqrt{25 - x^2} \, dx\). Use a substitution method: let \(u = 25 - x^2\), then compute \(du\) and rewrite the integral in terms of \(u\).

After evaluating the integral in terms of \(u\), substitute back the limits and solve for \(c\) by dividing both sides of the equation by the value of the integral to isolate \(c\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Probability Density Function (PDF)

A probability density function describes the likelihood of a continuous random variable taking on a particular value. For a function to be a valid PDF, it must be non-negative over its domain and its total integral over the specified interval must equal 1, representing the total probability.

Definite Integral for Normalization

To find the constant c that makes the function a valid PDF, you integrate the function over the given interval and set the integral equal to 1. Solving this equation for c ensures the area under the curve of the PDF equals 1, satisfying the normalization condition.

Integration Techniques Involving Square Roots

The integral involves a function with a square root term √(25 - x²), which suggests using substitution methods such as trigonometric substitution or recognizing it as a form related to the area of a circle segment. Mastery of these techniques is essential to evaluate the integral correctly.

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