Sol'n to PDE Integral: \frac{2b^2a}{3}

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Cyrus
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My PDE book does the following:

[tex]\int \phi_x^2 dx[/tex]

Where,

[tex]\phi_x = b-\frac{b}{a} |x|[/tex]

for [tex]|x|> a[/tex] and x=0 otherwise.

Strauss claims:

[tex]\int \phi_x^2 dx = ( \frac{b}{a} ) ^2 2a[/tex]

However, I think there is a mistake. It can be shown that:

[tex]\frac{-3a}{b}(b- \frac{b|x|}{a})^3[/tex] is a Soln. Evaluate this between 0<x<a and you get:

[tex]\frac{b^2 a}{3}[/tex]

Because the absolute value function is symmetric, its twice this value:

[tex]\frac{2b^2 a}{3}[/tex]

Unless I goofed, I think the book is in error.

*Note: Intergration is over the whole real line.
 
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I think you must mean |x|<a, right? Otherwise the integral isn't defined. I tried integrating (b-bx)^2 from 0 to a and I don't get anything close to either answer. Can you clarify?
 
O crap, its b-b/a|x| sorry. See above I fixed it.
 
Then I'm getting the same result as you.