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

[danoonez]

From

$$\int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1$$

show that

$$A = \sqrt {\frac {2} {L}$$



Here's what I did:

First bring $$A^2$$ out so that I have:

$$A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1$$

Then I use u subsitution for the integral:

$$\int sin^2 \frac {n \pi x} {L} dx$$

equals:

$$\int \frac {L} {n \pi} sin^2 u du$$

Pulling out the constants I get:

$$\frac {L} {n \pi} \int sin^2 u du$$

Which, using an integral table, equals:

$$(\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)$$

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

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

Plugging this back into the original equation I get:

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

Then solving for the limits of integration I get:

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

Then doing some algebra I get:

$$A^2 (\frac {n \pi L^2} {2 L n \pi} - \frac {sin 2 n \pi L^2} {4 L n \pi}) = 1$$

...then some scratch work...

$$sin \pi = 0$$

...so...

$$(A^2) (\frac {n \pi L^2} {2 L n \pi}) = 1$$

After more algebra I get:

$$(A^2) (\frac {L} {2}) = 1$$

So then:

$$A^2 = \frac {2} {L}$$

And finally:

$$A = \sqrt {\frac {2} {L}$$

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]

$$(\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)$$

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

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

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

Everything else looks OK.

 

[danoonez]

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

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.