The normalized solution to the Schrodinger equation for a particular potential is psi = 0 for x 0. What is the probability of finding a particle in this potential between x = a - 0.027a and x = a + 0.027a?

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.