Evaluate the improper integral or state that it diverges: integral from -inf to inf (2xe^-x) dx. Please help!
∫2x e^-x dx = -2 e^-x (x+1) = -2(x+1)/e^x
Now, e^x grows so much faster than x that the limit is zero as x -> ∞.
That can be justified by using lHospital's Rule, to show that the limit is the same as
-2/e^x -> 0
However, as x -> -∞ e^x grows very fast, so the limit is -∞
To evaluate the improper integral ∫-∞ to ∞ (2xe^(-x)) dx, we can use the technique of integration by parts.
Integration by parts is based on the following formula:
∫u dv = uv - ∫v du
Let's assign u = x and dv = 2e^(-x) dx. Taking the derivatives and integrals, we have:
du = dx
v = -2e^(-x)
Now, we can use the formula to solve the integral:
∫-∞ to ∞ (2xe^(-x)) dx = [-2xe^(-x)] - ∫-∞ to ∞ (-2e^(-x)) dx
Let's evaluate each term separately:
First term: [-2xe^(-x)]
Using the L'Hôpital's rule, we find that as x approaches -∞, (-x) / e^x approaches 0. Therefore, the limit of -2xe^(-x) as x goes to -∞ is 0.
Second term: ∫-∞ to ∞ (-2e^(-x)) dx
The integral of -2e^(-x) dx can be evaluated as follows:
∫-∞ to ∞ (-2e^(-x)) dx = [-2e^(-x)] evaluated from -∞ to ∞
= [-2e^(-∞)] - [-2e^(∞)]
As e^(-∞) approaches 0, and e^(∞) approaches ∞, we can conclude that the second term is undefined or divergent.
Therefore, the improper integral ∫-∞ to ∞ (2xe^(-x)) dx is divergent.
I hope this explanation helps you understand how to evaluate the improper integral and determine whether it converges or diverges. If you have any further questions, feel free to ask!