$E[X]$ derivation in Coupon Collecting Problem in Ross Probability Models In Ross Probability Models 11th edition on page 307, the following is written (in picture link). 
Coupon Collecting $E[X]$ derivation. $X=\max(X_j)$ is the time at which we obtain the collection of the coupons, and $X_{j}$ is independent exponential. I am not following why $E[X]=\int_{0}^{\infty} P(X \ge t)dt$?
This isn't specific to this particular random variable $X$. It's a general expression for the expected value of a non-negative random variable that you can derive from the corresponding integral with the density using integration by parts:

$$
E[X]=\int_0^\infty f_X(t)t\mathrm dt=\left[-P(X\ge t)t\right]_0^{\infty}+\int_0^\infty P(X\ge t)\mathrm dt\;.
$$
Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $X:\Omega\to\mathbb R$ a random variable with $\mathbb P(X\geqslant0)=1$. The map $(\omega,x)\mapsto \mathsf 1_{\{X(\omega)&gt;x\}}$ on $\Omega\times[0,\infty)$ is $\mathcal F\otimes\mathcal B(\mathbb R)$-measurable and is nonnegative. The measure $\mathbb P\otimes m$ ($m$ denoting Lebesgue measure) on $\Omega\times[0,\infty)$ is $\sigma$-finite, and hence by Tonelli's theorem the integral
$$
\iint_{\Omega\times[0,\infty)} \mathsf 1_{\{X(\omega)&gt;x\}} \ \mathsf d(\mathbb P\times x) $$
exists and is equal to the iterated integrals in the sequel. One has 
\begin{align}
\int_\Omega\int_{[0,\infty)}\mathsf 1_{\{X(\omega)&gt;x\}}\ \mathsf dx\ \mathsf d\mathbb P(\omega) &amp;= \int_\Omega\int_0^{X(\omega)}\ \mathsf dx\ \mathsf d\mathbb P(\omega)\\ &amp;= \int_\Omega X(\omega)\ \mathsf d\mathbb P(\omega)\\ &amp;= \mathbb E[X]
\end{align}
by definition of expectation, and further
\begin{align}
\int_{[0,\infty)}\int_\Omega \mathsf 1_{\{X(\omega)&gt;x\}}\ \mathsf d\mathbb P(\omega)\ \mathsf dx &amp;= \int_{[0,\infty)}\mathbb P(\omega\in\Omega:X(\omega)&gt;x)\ \mathsf dx\\
&amp;=\int_0^\infty \mathbb P(X&gt;x)\ \mathsf dx.
\end{align}
Hence,
$$
\mathbb E[X] = \int_0^\infty \mathbb P(X&gt;x)\ \mathsf dx.
$$