How do you integrate
(x-r)^2 (1/r) (e^(-x/r)) from the interval negative infinity to positive infinity?
To integrate the given expression, we can use the fact that the integral of a product of functions is equal to the product of the integrals of each individual function. Therefore, we will integrate each term separately.
Let's break down the integrand:
(x - r)^2 (1/r) (e^(-x/r))
To integrate (x - r)^2, we can expand it using the binomial theorem:
(x - r)^2 = x^2 - 2xr + r^2
Now, let's integrate each term:
∫ x^2 (1/r) (e^(-x/r)) dx
∫ (-2xr) (1/r) (e^(-x/r)) dx
∫ r^2 (1/r) (e^(-x/r)) dx
Integrating each term involves applying the power rule for integration, as well as the chain rule for the exponential function.
Let's start with the first term: ∫ x^2 (1/r) (e^(-x/r)) dx
We can rewrite this term as: (1/r) ∫ x^2 (e^(-x/r)) dx
To integrate x^2 (e^(-x/r)), we can use integration by parts. Integration by parts is a technique that involves splitting the integrand into two parts, one to differentiate and the other to integrate.
Let's define u = x^2 and dv = (1/r) (e^(-x/r)) dx. Therefore, du = 2x dx and v = ∫ (1/r) (e^(-x/r)) dx.
Using integration by parts:
∫ u dv = uv - ∫ v du
Applying the formula, we have:
(1/r) ∫ x^2 (e^(-x/r)) dx = (1/r) [-x^2 (r e^(-x/r)) - ∫ -2x (r e^(-x/r)) dx]
Simplifying this expression gives us:
∫ x^2 (1/r) (e^(-x/r)) dx = -x^2 e^(-x/r) + 2 ∫ x (e^(-x/r)) dx
Next, let's move on to the second term: ∫ (-2xr) (1/r) (e^(-x/r)) dx
Notice that (1/r) gets canceled out with (1/r), and we are left with:
-2 ∫ x (e^(-x/r)) dx
Finally, for the third term ∫ r^2 (1/r) (e^(-x/r)) dx:
The r in the numerator cancels out with (1/r) and simplifies to:
r ∫ (e^(-x/r)) dx
To integrate e^(-x/r), we can use the substitution method.
Let's substitute u = -x/r, which implies du = -dx/r.
Rearranging the equation, we have dx = -r du.
Now, let's rewrite the integral:
r ∫ (e^(-x/r)) dx = r ∫ (e^u) (-r du)
Simplifying this expression gives us:
- r^2 ∫ (e^u) du
The integral of e^u is simply e^u itself:
- r^2 e^u
Now, substituting back u = -x/r:
- r^2 e^(-x/r)
Now, we have all the individual integrals:
- x^2 e^(-x/r) + 2 ∫ x (e^(-x/r)) dx -2 ∫ x (e^(-x/r)) dx - r^2 e^(-x/r)
Combining these terms, we get:
- x^2 e^(-x/r) - r^2 e^(-x/r)
To find the integral from negative infinity to positive infinity, we need to compute the limits as x approaches negative and positive infinity.
As x approaches negative infinity, both x^2 and e^(-x/r) tend to infinity, so the first term (- x^2 e^(-x/r)) goes to negative infinity.
As x approaches positive infinity, both x^2 and e^(-x/r) tend to infinity, so the first term (- x^2 e^(-x/r)) still goes to negative infinity.
Therefore, the integral of the given expression from negative infinity to positive infinity is negative infinity.