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You're not understading:Originally posted by Tom

What you did wrong was take the answer off a calculator.

Do a u-substitution:

u=(π/L)x

du=(π/L)dx

then look up ∫sin^{2}(u)du in an integral table. Don't forget to evaluate it at each limit.

Let me give you all my work to alleaviate any confusion.

Show that A = (2/L)

^{1/2}

&psi(x) = A Sin(&pi x/L)

&psi

^{2}(x) = A

^{2}Sin

^{2}(&pi x/L)

[inte]

_{0}

^{L}&psi

^{2}dx = 1

A

^{2}[inte]

_{0}

^{L}Sin

^{2}(&pi x/L) dx = 1

Actually...

I forgot to resubsitute...

BTW: I only use a calculator/computer to check my work.

Thanks...

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