Problem

The Fresnel function $ S $ was defined in Example…

06:52

Problem 70 Hard Difficulty

The error function $$ \displaystyle \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} \, dt $$ is used in probability, statistics, and engineering.
(a) Show that $ \displaystyle \int^b_a e^{-t^2} \, dt = \frac{1}{2} \sqrt{\pi} [ \text{erf}(b) - \text{erf}(a) ] $.
(b) Show that the function $ y = e^{x^2} \text{erf}(x) $ satisfies the differential equation $ y' = 2xy + 2/\sqrt{\pi} $.

Answer

a.Click to see the solution.
b.$$2 x y+\frac{2}{\sqrt{\pi}}$$

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Video Transcript

It's like every were given the error function which is used in probability statistics and engineering. The error function is defined to be Two over the square root of pi times the integral from 0 to X of each. The negative T squared D T is I'm afraid of action. In part, they were asked to show that the integral from A to B of each of the negative T squared DT can be written in terms of error functions. So the integral from A to B. Eat a negative t squared DT. Well, who wants to the kids? If you wanted those two fellows for instance, Big favorite system Harleys. We can rewrite this as the integral from zero to be of E to the negative T squared DT plus the integral from a +20 of E to the negative T squared Bt. Using integral properties, we can rewrite this as everything I have Integral from 0 to be. In fact, I'll write this as route high over to here times two of a root pie. And the integral from zero to be of E. To the negative T square D. T minus two of the Root Pie. Times. In the girl from zero to a. Of each of the negative T squared DT. The notice that by definition this is rude pi over two times error function evaluated at B minus the error function evaluated at a. Yeah, this is what we wanted to show for part egg. Mhm. Yeah. Then in part B we're trying to lose weight from china for us to show that the function Y equals E. To the X squared times the error function evaluated at X satisfies the differential equation. Why prime equals two X one Plus two over the square root of Pi. Well, all we have to do is simply plugs our function into this differential equation on both sides. So first let's find the left hand side. Wide prime. Buy the product rule. This is a derivative of E. V. X squared which is uh two X times E. To the X squared times the error function evaluated at X plus E. To the X squared times the derivative of the air function which this is going to be to overrule pie. And then by the fundamental theorem of calculus, the derivative of the error function is two of the route high times each of the negative X squared. Since the function E. To the negative T squared is continuous for all values of T. Yeah. Mhm. Yeah. As he Hard that's hard. eight are they hard up? Here's the thing bigger animals actually lost. And sure now that's the left hand side. The only way to see used. Do you do you do? I'm up party. Yeah. So you know that saying the house you can join now joined right now by partly my sister. So actually could simplify first and this is equal to two X times E. To the X squared air function of X plus. And this is simply two of her route pie Times each of the zero which is the same as one. Now we can rearrange since error function is Y over E. To the X squared. This is two X times not during not during the ship wow. Where we're just either the X squared error function. X. This is why This is two x. Y Plus two Over Group I. No and this is the right hand side of the equation. Therefore the function is a solution to this differential equation. That's probably enterprise. No that's not. Oh it's coming. Oh I can tell you.

Chris T.

Ohio State University

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Problem

The Fresnel function $ S $ was defined in Example…

Problem 70 Hard Difficulty

Answer

a.Click to see the solution.
b.$$2 x y+\frac{2}{\sqrt{\pi}}$$

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Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

a.Click to see the solution.b.$$2 x y+\frac{2}{\sqrt{\pi}}$$

Calculus: Early Transcendentals

Grace H.

Catherine R.

Kayleah T.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Grace H.

Catherine R.

Kayleah T.

Kristen K.

a.Click to see the solution.
b.$$2 x y+\frac{2}{\sqrt{\pi}}$$