# Expectation Value

## 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)

## The Attempt at a Solution

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

$$\int_0^a A^2 \frac{ x^3 }{ a^2} = A^2 \frac{ a^2 }{4} = 3/4 a^2/b$$

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

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

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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Dick
Homework Helper
The antiderivative of the x(b-x)^2 does NOT work. Try it again. How did you get that form anyway?

I used integration by parts,

u = x
du = dx

dv = (b-x)^2
v = -(b-x)^3/3

vu - int(v du)

$$\frac{-(b-x)^3 x}{3} - \frac{(b-x)^4}{12}$$

so
$$A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = \frac{A^2}{(b-a)^2} * (\frac{-x(b-x)^3}{4} - \frac{(b-x)^4}{12})$$
I ran through it again but I can't seem to see what I am doing incorrectly.

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The 0 to a integration seems to be correct.

The a to b integration seems little bit weird. The approach that you did by using the integration by part can lead you to the answer.

$$A^2 \int_a^b \frac{x(b-x)^2}{(b-a)^2} dx = \frac{A^2}{(b-a)^2} * (\frac{-x(b-x)^3}{3} - \frac{(b-x)^4}{12})$$

$$= \frac{3}{b} (\frac{a(b-a)}{3}+\frac{(b-a)^2}{12})$$

Now you can get <x> value by doing:

$$<x>=\frac{3a^2}{4b}+\frac{4a(b-a)}{4b}+\frac{(b-a)^2}{4b}=\frac{2a+b}{4}$$

Pretty much a straight forward process.

Hope it helped =)

Dick