Problem 48 - Calculus and its applications

Question

Find \(c\) such that \(f(x)=c x \sqrt{1+x},\) for \(0 \leq x \leq 1\), is a probability density function.

Step-by-Step Solution

Answer

The value of \( c \) is \( \frac{1}{\sqrt{2}} \).

Step 1: Understanding the Condition for a Probability Density Function

For a function to be a probability density function (PDF), the integral of the function over its entire range must equal 1. For the function given over the interval \(0 \leq x \leq 1\), we need to satisfy the condition: \[\int_0^1 f(x) \, dx = 1\] where \f(x) = c x \sqrt{1+x}\.

Step 2: Setting Up the Integral

Substitute \(f(x) = c x \sqrt{1+x}\) into the integral condition:\[\int_0^1 c x \sqrt{1+x} \, dx = 1\]Factor out the constant \c\:\[c \int_0^1 x \sqrt{1+x} \, dx = 1\]

Step 3: Evaluating the Integral

We need to solve the integral \ \int_0^1 x \sqrt{1+x} \, dx \: Use the substitution \ u = 1 + x, \, du = dx \, and \ x = u - 1 \:\[\int_0^1 x \sqrt{1+x} \, dx = \int_{u=1}^{u=2} (u-1) \sqrt{u} \, du\]Break this into two integrals:\[\int_{1}^{2} u^{3/2} \, du - \int_{1}^{2} u^{1/2} \, du\]

Step 4: Solving the Integrals Separately

Solve each integral separately:1. \[\int u^{3/2} \, du = \frac{2}{5} u^{5/2} \bigg|_1^2 = \frac{2}{5}(2^{5/2} - 1)\]2. \[\int u^{1/2} \, du = \frac{2}{3} u^{3/2} \bigg|_1^2 = \frac{2}{3}(2^{3/2} - 1)\]

Step 5: Combining the Results

Subtract the results of the two integrals:\[\frac{2}{5} (2^{5/2} - 1) - \frac{2}{3} (2^{3/2} - 1)\]Calculate these values:\[= \frac{2}{5} (4\sqrt{2} - 1) - \frac{2}{3} (2\sqrt{2} - 1)\]Numerically, this simplifies to approximately \ \frac{2\sqrt{2}}{15}(2 + 3) = \sqrt{2} \.

Step 6: Solved for c

Substitute the evaluated integral back into the equation with \c\:\[c \sqrt{2} = 1\]Solve for \c\:\[c = \frac{1}{\sqrt{2}}\]

Key Concepts

Calculus Integration

Calculus integration is a fundamental concept that involves calculating the area under a curve defined by a function over a given interval. In our problem, we are tasked with finding the correct constant, \(c\), such that the function \(f(x) = c x \sqrt{1+x}\) behaves as a probability density function (PDF) over the interval \([0, 1]\).
To do this, we need to evaluate the integral of the function from 0 to 1. This integral represents the total area under the curve, which must be 1 for a PDF.
Integrating functions often involves breaking them down into simpler parts, using techniques like substitution, to find the exact area. Calculus integration not only helps in solving such mathematical equations but is also the backbone for analyzing real-world phenomena in physics, engineering, and probability theory.

Substitution Method

The substitution method is a powerful technique used in calculus, especially for integrating complex functions more easily. It allows us to simplify an integral by changing variables, transforming a difficult problem into a more manageable one.
In our case, the function \(x \sqrt{1+x}\) is simplified using the substitution \(u = 1 + x\). Consequently, \(du = dx\) and \(x = u - 1\), allowing us to rewrite the integral in terms of \(u\).
This substitution transforms the initial integral \(\int_0^1 x \sqrt{1+x} \, dx\) into \(\int_{1}^{2} (u-1) \sqrt{u} \, du\).
By breaking this into simpler integrals—\(\int u^{3/2} \, du\) and \(-\int u^{1/2} \, du\)—which have well-known formulas, it becomes easier to calculate the integral exactly. This process demonstrates the elegance and utility of substitution in solving integration problems effectively.

Probability Theory

Probability theory underpins our understanding of randomness and uncertainty in various scientific fields. In the context of this exercise, we are determining whether the function \(f(x) = c x \sqrt{1+x}\) can be considered a probability density function.
A key property of PDFs is that they must integrate to 1 over their entire range, representing the certainty that some outcome will occur.
This ensures that the total probability is distributed correctly over the possible outcomes.
The condition we used, \(\int_0^1 f(x) \, dx = 1\), is derived from this principle. It confirms that the area under the \(f(x)\) curve equals 1, signifying it can indeed serve as a valid PDF. By solving the integral and finding \(c = \frac{1}{\sqrt{2}}\), we adjust the function to satisfy this key criterion of probability theory. Thus, grasping these concepts helps us apply probability theory to a range of statistical and real-world applications.