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[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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# Homework Help: Please check my integration

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