The error function erf(«) = [e*at is used in probability, statistics, and engineering. (a) Show that e- dt =VTlerf(b) – erf(a)]. (1) erf(x) = dt dt= erf(x) (2) By this property of definite integrals, e dt = dt + dt (1) and (2) = dt dt - dt erf(b) – erf a - ert a (b) Show that the function y = eerf(x) satisfies the differential equation y' = 2xy + 2/VT. y = e*erf(x) = y' = erf(x) + e*erf (x) 2x [by FTC1]
The error function erf(«) = [e*at is used in probability, statistics, and engineering. (a) Show that e- dt =VTlerf(b) – erf(a)]. (1) erf(x) = dt dt= erf(x) (2) By this property of definite integrals, e dt = dt + dt (1) and (2) = dt dt - dt erf(b) – erf a - ert a (b) Show that the function y = eerf(x) satisfies the differential equation y' = 2xy + 2/VT. y = e*erf(x) = y' = erf(x) + e*erf (x) 2x [by FTC1]
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### The Error Function
The error function, denoted as \( \text{erf}(x) \), is defined by the integral:
\[ \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt \]
This function is commonly used in the fields of probability, statistics, and engineering.
### Proofs and Properties
#### (a) Show that:
\[ \int_{a}^{b} e^{-t^2} dt = \frac{1}{2} \sqrt{\pi} [\text{erf}(b) - \text{erf}(a)] \]
1. Given:
\[ \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt \]
We need to show:
\[ \int_{0}^{x} e^{-t^2} dt = \text{erf}(x) \]
However, this should be corrected to:
\[ \int_{0}^{x} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} \text{erf}(x). \]
2. By using the property of definite integrals:
\[ \int_{a}^{b} e^{-t^2} dt = \int_{0}^{b} e^{-t^2} dt - \int_{0}^{a} e^{-t^2} dt \]
Combining (1) and (2), we get:
\[ \int_{a}^{b} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} \left[\text{erf}(b) - \text{erf}(a)\right]. \]
#### (b) Verification of Differential Equation
Show that the function \( y = e^{x^2} \text{erf}(x) \) satisfies the differential equation \( y' = 2xy + \frac{2}{\sqrt{\pi}} \).
Given:
\[ y = e^{x^2} \text{erf}(x) \]
Differentiate \( y \) with respect to \( x \):
\[ y' = \frac{d}{dx} \left(e^{x^2} \text{erf}(x)\right) \]
Using product rule](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F75ba17ec-6252-468f-8aa7-ac5f44232056%2Fd9cd6872-ed19-4e6f-95e9-febdf028f227%2Fbcy5a0q_processed.png&w=3840&q=75)
Transcribed Image Text:### The Error Function
The error function, denoted as \( \text{erf}(x) \), is defined by the integral:
\[ \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt \]
This function is commonly used in the fields of probability, statistics, and engineering.
### Proofs and Properties
#### (a) Show that:
\[ \int_{a}^{b} e^{-t^2} dt = \frac{1}{2} \sqrt{\pi} [\text{erf}(b) - \text{erf}(a)] \]
1. Given:
\[ \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt \]
We need to show:
\[ \int_{0}^{x} e^{-t^2} dt = \text{erf}(x) \]
However, this should be corrected to:
\[ \int_{0}^{x} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} \text{erf}(x). \]
2. By using the property of definite integrals:
\[ \int_{a}^{b} e^{-t^2} dt = \int_{0}^{b} e^{-t^2} dt - \int_{0}^{a} e^{-t^2} dt \]
Combining (1) and (2), we get:
\[ \int_{a}^{b} e^{-t^2} dt = \frac{\sqrt{\pi}}{2} \left[\text{erf}(b) - \text{erf}(a)\right]. \]
#### (b) Verification of Differential Equation
Show that the function \( y = e^{x^2} \text{erf}(x) \) satisfies the differential equation \( y' = 2xy + \frac{2}{\sqrt{\pi}} \).
Given:
\[ y = e^{x^2} \text{erf}(x) \]
Differentiate \( y \) with respect to \( x \):
\[ y' = \frac{d}{dx} \left(e^{x^2} \text{erf}(x)\right) \]
Using product rule
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