The time it takes for a lightbulb to burn out is exponentially distributed with mean $u$ which is a random variable. Asssume that $u$ is distributed with density
$$f(x)=\frac{8}{x^3}$$
for $x \in [2, \infty)$.
What is $P(\text{lightbulb is burned out by time } 7)=P(E)$?
$$P(E)=1-P(E^c)=1-\int e^{-7x}\frac{8}{x^3}dx$$
Where does the $-7x$ come from?
The conditional lifetime random variable for a fixed mean $\mu$ is given by $$T \mid \mu \sim \operatorname{Exponential}(\mu)$$ where $$f_{\mu}(x) = 8x^{-3}, \quad x \ge 2.$$ So in particular, recall that the CDF of an exponential distribution is $$\Pr[T \le t \mid \mu] = F_{T \mid \mu}(t) = 1 - e^{-t/\mu}, \quad t \ge 0.$$ Thus by the law of total probability $$\Pr[T \le 7] = \int_{x = 2}^\infty \Pr[T \le 7 \mid \mu = x] f_\mu(x) \, dx = \int_{x=2}^\infty (1 - e^{-7/x}) \cdot 8x^{-3} \, dx.$$ Note this is equivalent to what you wrote in your question, since we know that $$\int_{x=2}^\infty f_{\mu}(x) \, dx = 1.$$