Expected Value of the maximum of two exponentially distributed random variables

The minimum of two independent exponential random variables with parameters $\lambda$ and $\eta$ is also exponential with parameter $\lambda+\eta$.

Also $\mathbb E\big[\min(X_1,X_2)+\max(X_1,X_2)\big]=\mathbb E\big[X_1+X_2\big]=\frac{1}{\lambda}+\frac{1}{\eta}$. Because $\mathbb E\big[\min(X_1,X_2)\big]=\frac{1}{\lambda+\eta}$, we get $\mathbb E\big[\max(X_1,X_2)\big]=\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}.$

Let $V=\max\{X,Y\}$. Then $$\mathbb{P}(V\leq t)=\mathbb{P}(X\leq t,Y\leq t)=\mathbb{P}(X\leq t)\mathbb{P}(Y\leq t).$$ Now find $f_V(t)$ and then $\int_{-\infty}^{+\infty}tf_V(t)dt$, which should be $\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}$.

The sample $(X,Y)$ have a density given by $f_X(x)f_Y(y)$ since $X$ and $Y$ are independent. You have to compute $$\iint_{\Bbb R^2}\max\{x,y\}f_X(x)f_Y(y)dxdy.$$ Cut this integral in two parts.