Prove A = √(2/L) using Integration - Step by Step Guide

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Homework Help Overview

The discussion revolves around proving the equation A = √(2/L) through integration, specifically focusing on the integral of sin²(nπx/L) over the interval from 0 to L. The original poster presents their steps in manipulating the integral to derive the expression for A.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to simplify the integral by using substitution and algebraic manipulation. Some participants question the handling of terms during the substitution process, particularly regarding the sine function. Others suggest clarifying the notation used in the expressions.

Discussion Status

Participants are actively engaging with the original poster's steps, providing feedback on notation and potential typos. There is a collaborative effort to ensure clarity and correctness in the mathematical expressions presented, with no explicit consensus reached on the final outcome.

Contextual Notes

There are mentions of potential typos and notation issues that may affect the clarity of the presented solution. The discussion reflects a focus on ensuring the integrity of the mathematical reasoning without resolving the original problem completely.

danoonez
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From

[tex] \int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1[/tex]

show that

[tex] A = \sqrt {\frac {2} {L}[/tex]



Here's what I did:

First bring [tex]A^2[/tex] out so that I have:

[tex] A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1[/tex]

Then I use u subsitution for the integral:

[tex] \int sin^2 \frac {n \pi x} {L} dx[/tex]

equals:

[tex] \int \frac {L} {n \pi} sin^2 u du[/tex]

Pulling out the constants I get:

[tex] \frac {L} {n \pi} \int sin^2 u du[/tex]

Which, using an integral table, equals:

[tex] (\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)[/tex]

Plugging u back in and putting in my limits of integration I get my definite integral:

[tex] (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L}[/tex]

Plugging this back into the original equation I get:

[tex] A^2 (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L} = 1[/tex]

Then solving for the limits of integration I get:

[tex] A^2 (\frac {L} {n \pi}) \left[ (\frac {n \pi L} {2L} - \frac {sin 2 n \pi L} {4 L}) - (0 - 0) \right] = 1[/tex]

Then doing some algebra I get:

[tex] A^2 (\frac {n \pi L^2} {2 L n \pi} - \frac {sin 2 n \pi L^2} {4 L n \pi}) = 1[/tex]

...then some scratch work...

[tex] sin \pi = 0[/tex]

...so...

[tex] (A^2) (\frac {n \pi L^2} {2 L n \pi}) = 1[/tex]

After more algebra I get:

[tex] (A^2) (\frac {L} {2}) = 1[/tex]

So then:

[tex] A^2 = \frac {2} {L}[/tex]

And finally:

[tex] A = \sqrt {\frac {2} {L}[/tex]

Q.E.D.


How does it look.
Sorry if it seems messy; this was my first time using LaTex.
 
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There's a step around plugging n*pi*x/L back in for u where inexplicably the L in n*pi*x/L was pulled out of the sine function. Was this a typo?
 
danoonez said:
[tex] (\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)[/tex]

Plugging u back in and putting in my limits of integration I get my definite integral:

[tex] (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L}[/tex]

You should really put parentheses around the variables in the sine function, i.e.
[tex] \Big(\frac {L} {n \pi}\Big) \left[ \frac {n \pi x} {2L} - \frac {\sin (2 n \pi x/L)} {4} \right]_{0}^{L}[/tex]

Everything else looks OK.
 
e(ho0n3 said:
You should really put parentheses around the variables in the sine function, i.e.
[tex] \Big(\frac {L} {n \pi}\Big) \left[ \frac {n \pi x} {2L} - \frac {\sin (2 n \pi x/L)} {4} \right]_{0}^{L}[/tex]

Everything else looks OK.


Thank you. I will do that for the paper I hand in.


vsage: I can't tell which step you mean. Could you clarify? B/c it may not have been a typo.


Anybody else see any mistakes?
 
I was referring to what e(ho0n3 pointed out. You seem to account for the typo though later so it looks great to me.
 
Excellent. Thanks for looking.
 

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