Problem 8 - Calculus and its applications

Question

Verify Property 2 of the definition of a probability density function over the given interval. $$ f(x)=\frac{1}{e-1} e^{x}, \quad[0,1] $$

Step-by-Step Solution

Answer

The integral of the PDF \( f(x) \) over \([0, 1]\) equals 1, confirming it is valid.

Step 1: Identify Property 2 of PDF

The second property of a probability density function (PDF) states that the integral of the function over the given interval should equal 1. In our case, we need to verify that the integral of \( f(x) = \frac{1}{e-1} e^x \) over \([0, 1]\) is equal to 1.

Step 2: Set Up the Integral

The next step is to set up the integral of \( f(x) \) over the interval \([0, 1]\). We have: \[ \int_{0}^{1} \frac{1}{e-1} e^x \, dx \]

Step 3: Compute the Antiderivative

To integrate \( \frac{1}{e-1} e^x \), find the antiderivative of \( e^x \). The antiderivative of \( e^x \) is \( e^x \). Therefore, the antiderivative of \( \frac{1}{e-1} e^x \) is \( \frac{1}{e-1} e^x \).

Step 4: Evaluate the Definite Integral

Using the Fundamental Theorem of Calculus, evaluate the integral: \[ \left[ \frac{1}{e-1} e^x \right]_{0}^{1} = \frac{1}{e-1} e^1 - \frac{1}{e-1} e^0 \] Simplifying this gives:\[ \frac{1}{e-1} e - \frac{1}{e-1} \cdot 1 = \frac{1}{e-1} (e - 1) \]

Step 5: Verify Result Equals 1

Simplify the expression from Step 4: \[ \frac{1}{e-1} (e - 1) = 1 \] This confirms that the integral of the PDF over the interval \([0, 1]\) is equal to 1, satisfying Property 2.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals. It's a way to find the total value from a rate of change. In the context of probability density functions (PDF), integral calculus is used to determine the probability over an interval. By integrating a function, we calculate the area under its curve between two points on the x-axis. This area represents the probability of the random variable falling within that interval.

For example, consider the function given, \( f(x) = \frac{1}{e-1} e^x \), over the interval \([0, 1]\). To find the probability that a random variable falls between 0 and 1, we integrate \( f(x) \) from 0 to 1.

This process involves:

• Identifying the function to be integrated.
• Determining the limits of integration, which are 0 and 1 in our problem.
• Computing the integral to find the total area under the curve.

In our solution, after performing these steps, we confirmed that the integral equals 1, signifying it is a valid PDF on this interval.

Antiderivative

Finding an antiderivative is a crucial step in solving integral calculus problems. An antiderivative, also known as a primitive, is a function whose derivative is the given function. In other words, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f \).

When calculating definite integrals, the antiderivative helps determine the accumulated value across an interval. In this exercise, we've identified that the antiderivative of \( e^x \) is \( e^x \) itself. Therefore, the antiderivative of \( \frac{1}{e-1} e^x \) becomes \( \frac{1}{e-1} e^x \).

Using the antiderivative:

• Substitute the upper limit of the interval into the antiderivative.
• Subtract the result of substituting the lower limit from this value.

This process allows us to find the exact area under the curve, confirming the function's validity as a PDF.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, showing that they are essentially inverse processes. This theorem has two parts. The first deals with derivatives of integrals, and the second allows us to evaluate definite integrals using antiderivatives.

For our PDF verification, we use the second part of this theorem. It states that if \( F \) is an antiderivative of \( f \) over an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) equals \( F(b) - F(a) \).

Here's how we applied it:

• We found the antiderivative of the function, \( F(x) = \frac{1}{e-1} e^x \).
• Then, we evaluated \( F(x) \) at the bounds 1 and 0.
• Finally, subtracting these results allowed us to confirm the integral amounts to 1.

The Fundamental Theorem of Calculus streamlines the process of finding total values from rates of change, essential in confirming properties of probability density functions.