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## Homework Statement

Find the expectation value <x> if:

from 0 <= x <= a, psi = A x/a

from a <= x <= b, psi = A(b-x)/(b-a)

Normalizing gives me that A = sqrt(3/b) (verified correct)

## Homework Equations

## The Attempt at a Solution

[tex]<x> = \int_0^b x \psi^2 dx = \int_0^a \frac{A^2 x^3}{a^2}dx + \int_a^b A^2 \frac{x (b-x)^2 }{(b-a)^2} dx[/tex]

[tex]\int_0^a A^2 \frac{ x^3 }{ a^2} = A^2 \frac{ a^2 }{4} = 3/4 a^2/b[/tex]

[tex]A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = -A^2 \frac{x (b-x)^3}{(b-a)^2} - A^2 \frac{(b-x)^4}{12(b-a)^2}[/tex] eval from b to a

[tex]<x> = \frac{3a^2}{4b } + \frac{+ 3(b-a)a}{b} + \frac{3(b-a)^2}{12b} = \frac{-8a^2 + b^2 + 10ab}{4b}[/tex]

The answer is supposed to be (2a+b)/4... Does anyone see where I've gone wrong? I have double checked the antideriv. for the x(b-x)^2 integral and it seems to work.

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