The normalized solution to the Schrodinger equation for a particular potential is `psi` = 0 for x 0....

Hello!

The probability of being between some c and d is `int_c^d |Psi(x)|^2 dx.` Therefore the probability in question is

`int_(a-0.027a)^(a+0.027a) |Psi(x)|^2 dx =int_(a-0.027a)^(a+0.027a) 4/a^3 x^2 e^(-(2x)/a) dx`

(`a` must be positive, so bounds of integration are also positive).

To compute the indefinite integral of `x^2 e^(-(2x)/a)` we can use integration by parts twice: differentiate `x^2` and then `x` and integrate the exponent. Let's perform this:

`int x^2 e^(-(2x)/a) dx = |u = x^2, dv...

Hello!

The probability of being between some c and d is `int_c^d |Psi(x)|^2 dx.` Therefore the probability in question is

`int_(a-0.027a)^(a+0.027a) |Psi(x)|^2 dx =int_(a-0.027a)^(a+0.027a) 4/a^3 x^2 e^(-(2x)/a) dx`

(`a` must be positive, so bounds of integration are also positive).

To compute the indefinite integral of `x^2 e^(-(2x)/a)` we can use integration by parts twice: differentiate `x^2` and then `x` and integrate the exponent. Let's perform this:

`int x^2 e^(-(2x)/a) dx = |u = x^2, dv = e^(-(2x)/a) dx, du = 2x dx, v = -a/2e^(-(2x)/a)| =`

`= -a/2 x^2e^(-(2x)/a) + a/2 int (2xe^(-(2x)/a)) dx.`

Then `u=x, dv =e^(-(2x)/a) dx, du = dx, v = -a/2 e^(-(2x)/a),` and the remaining integral is equal to

`-a/2 xe^(-(2x)/a) + a/2 inte^(-(2x)/a) dx =-xe^(-(2x)/a) - a/2 * a/2e^(-(2x)/a).`

So the total indefinite integral is equal to `-a/4 e^(-(2x)/a)(a^2+2ax+2x^2)+C,` and the probability is

`-4/a^3*a/4 (e^(-2(1+0.027))(a^2+2a^2(1+0.027)+2a^2(1+0.027)^2)-`

`- e^(-2(1-0.027))(a^2+2a^2(1-0.027)+2a^2(1-0.027)^2))).`

`a` vanishes and remains

`e^(-2(1-0.027))(1+2(1-0.027)+2(1-0.027)^2) -`

`- e^(-2(1+0.027))(1+2(1+0.027)+2(1+0.027)^2) approx 0.0292.`

This is the answer.