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

[danoonez]

From

<br /> \int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1<br />

show that

<br /> A = \sqrt {\frac {2} {L}<br />



Here's what I did:

First bring A^2 out so that I have:

<br /> A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1<br />

Then I use u subsitution for the integral:

<br /> \int sin^2 \frac {n \pi x} {L} dx<br />

equals:

<br /> \int \frac {L} {n \pi} sin^2 u du<br />

Pulling out the constants I get:

<br /> \frac {L} {n \pi} \int sin^2 u du<br />

Which, using an integral table, equals:

<br /> (\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)<br />

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

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

Plugging this back into the original equation I get:

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

Then solving for the limits of integration I get:

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

Then doing some algebra I get:

<br /> A^2 (\frac {n \pi L^2} {2 L n \pi} - \frac {sin 2 n \pi L^2} {4 L n \pi}) = 1<br />

...then some scratch work...

<br /> sin \pi = 0<br />

...so...

<br /> (A^2) (\frac {n \pi L^2} {2 L n \pi}) = 1<br />

After more algebra I get:

<br /> (A^2) (\frac {L} {2}) = 1<br />

So then:

<br /> A^2 = \frac {2} {L}<br />

And finally:

<br /> A = \sqrt {\frac {2} {L}<br />

Q.E.D.


How does it look.
Sorry if it seems messy; this was my first time using LaTex.

 

[vsage]

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?

 

[e(ho0n3]

<br /> (\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)<br />

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

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

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

Everything else looks OK.

 

[danoonez]

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

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?

 

[vsage]

I was referring to what e(ho0n3 pointed out. You seem to account for the typo though later so it looks great to me.

 

[danoonez]

Excellent. Thanks for looking.