# Probability of $X_1 > 2X_2$, where $X_1$ and $X_2$ are independent random variables with exponential distribution functions

I have two random variables, $$X_1$$ and $$X_2$$, both of which have exponential distributions, and $$\lambda_1 = 1$$, $$\lambda_2 = 2$$. The question is then, what is the probability $$P(X_1 > 2X_2)$$. So I have two distribution functions:

$$f(x_1) = e^{-x_1}$$

$$f(x_2) = 2e^{-2x_2}$$

Looking at the integral I get

$$\int \int_{x_1 > x_2}f_1(x_1)f_2(x_2) dx_1 dx_2$$

Combining this all I come to

$$\int_0^\infty \int_{2x_2}^\infty 2e^{-(x_1 + x_2)}$$

This is my concern. Is the lower limit on my second integral actually $$2x_2$$ or should it be $$2x_1$$. To me $$2x_2$$ makes more sense, and leads to a result of $$\frac{1}{2}$$. However having a lower limit of $$2x_1$$ leads to a result of $$\frac{1}{5}$$ and I have seen this elsewhere. Can anybody perhaps shed some light on my issue and help me out with my confusion?

$$\int_0^\infty \int_{2x_2}^\infty 2e^{-(x_1+x_2)}\, dx_1 \, dx_2$$
Let $$x_2$$ takes value from $$0$$ to $$\infty$$. After which, fixing $$x_2$$, what are the values that $$x_1$$ can take to satisfy the domain of interest. We want $$x_1 >2x_2$$, hence the lower limit should be $$2x_2$$.