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\begin{align*} & \int_0^{\infty} \lambda e^{-\lambda x} dx\\ & = \int_0^{\infty} e^{-u} du\\ & = -e^{-u} \big |_0^{\infty} = 1 \end{align*}
where we used the substitution
\(u = \lambda x\text{.}\)
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