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Take the random variable $X$ with distribution $$f_X(x) = 6e^{-2x}, -k \leq x \leq k.$$ I want to find the distribution of $X+X^2.$ I believe that the first step of this is to figure out the $k$ that makes $f_X$ a legal PDF. We see that $$\int_{-k}^k 6e^{-2x} dx = 6 \int_{-k}^{k} e^{-2x} dx.$$ we see $u = -2x \implies du = -2dx$, and thus $$=-3 \int_{-k}^{k} e^u du = -3 ( e^{-2x} \mid_{-k}^k) = -3(e^{-2k} - e^{2k}) = 1$$ $$\implies \frac{1}{-3} = e^{-2k} - e^{2k},$$

and I am having some issues finding the $k$ that makes this correct. Any recommendations?

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You are correct all the way through, and just need to consider using a hyperbolic function identity.

Recall that $\sinh(x) = \frac{e^x-e^{-x}}{2}$ and note that

$$6\int_{-k}^k \mathrm{e}^{-2x}\,\mathrm{d}x = -6\left(\frac{e^{-2k}-e^{2k}}{2}\right)= 6\left(\frac{e^{2k}-e^{-2k}}{2}\right) = 6\sinh(2k)$$

Now, simply solve the equation $6\sinh(2k) = 1$ for $k$.

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