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Give a convincing argument that f(x)dx=1

Give a convincing argument that f(x)dx=1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I need part C. It has to show that the integral from -infinity to -b and the integral from infinity to b equal 0 and the integral of -b to b equals 1 to prove the question. Please show all work!!!!
Certainly! Here's a transcription and explanation suitable for an educational website. --- ### Problem Statement: Evaluate the integral \[ \int_{-n}^{n} f(x) \, dx \] for \( n = 1, 2, \) and \( 3 \). ### Part C: Convincing Argument Provide a convincing argument that \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] #### Hint: 1. Demonstrate that \( 0 < f(x) < e^{-x^2/2} \) for \( x > 1 \). 2. Show that for \( b > 1 \), \[ \int_{b}^{\infty} e^{-x^2/2} \, dx \to 0 \] as \( b \to \infty \). --- This exercise involves evaluating definite integrals for specific values of \( n \) and proving a condition for an improper integral. The hint suggests comparing \( f(x) \) with the Gaussian function \( e^{-x^2/2} \) to establish integrability constraints as \( x \) approaches infinity.
Transcribed Image Text:Certainly! Here's a transcription and explanation suitable for an educational website. --- ### Problem Statement: Evaluate the integral \[ \int_{-n}^{n} f(x) \, dx \] for \( n = 1, 2, \) and \( 3 \). ### Part C: Convincing Argument Provide a convincing argument that \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] #### Hint: 1. Demonstrate that \( 0 < f(x) < e^{-x^2/2} \) for \( x > 1 \). 2. Show that for \( b > 1 \), \[ \int_{b}^{\infty} e^{-x^2/2} \, dx \to 0 \] as \( b \to \infty \). --- This exercise involves evaluating definite integrals for specific values of \( n \) and proving a condition for an improper integral. The hint suggests comparing \( f(x) \) with the Gaussian function \( e^{-x^2/2} \) to establish integrability constraints as \( x \) approaches infinity.
**Normal Probability Distribution** The function \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x-\mu}{\sigma}\right)^2} \] is called the *normal probability density function* with the mean \(\mu\) and standard deviation \(\sigma\). The number \(\mu\) tells where the distribution is centered, and \(\sigma\) measures the “scatter” around the mean. From the theory of probability, it is known that \[ \int_{-\infty}^{\infty} f(x) \, dx = 1. \] In what follows, let \(\mu = 0\) and \(\sigma = 1\).
Transcribed Image Text:**Normal Probability Distribution** The function \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x-\mu}{\sigma}\right)^2} \] is called the *normal probability density function* with the mean \(\mu\) and standard deviation \(\sigma\). The number \(\mu\) tells where the distribution is centered, and \(\sigma\) measures the “scatter” around the mean. From the theory of probability, it is known that \[ \int_{-\infty}^{\infty} f(x) \, dx = 1. \] In what follows, let \(\mu = 0\) and \(\sigma = 1\).
Expert Solution
Step 1: step 1

We start with the definition of the normal probability density function:

f(x)=12πσ2e−(x−μ)2/2σ2

where is the mean of the distribution and is the variance.

We then substitute this function into the integral:

∫−bbf(x)dx=∫−bb12πσ2e−(x−μ)2/2σ2dx

We can now use the following change of variables:

z=x−μσ

This change of variables transforms the integral into the following:

∫−bbf(x)dx=∫−(b−μ)/σ(b−μ)/σ12πe−z2/2dz

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