The value for $E \left[ \max \left( L_1, L_2 \right) \right]$ is computed in the following way. First, the distribution of the maximum of two identically independently distribued random variable $L_1$ and $L_2$ is given by $2 f \left( \ell \right) F \left( \ell \right)$ where $f \left( \ell \right)$ is the density and $F \left( \ell \right)$ is the cumulative distribution function. This is well known, you could find the formula here. It is not difficult to derive: \begin{eqnarray*} \Pr \left[ \max \left( L_1, L_2 \right) \leqslant \ell \right] & = & \Pr \left[ \left\{ L_1 \leqslant \ell \right\} \cap \left\{ L_2 \leqslant \ell \right\} \right]\\ & = & \Pr \left[ L_1 \leqslant \ell \right] \Pr \left[ L_2 \leqslant \ell \right]\\ & = & F \left( \ell \right)^2 \end{eqnarray*} Taking derivative gives the density $2 f \left( \ell \right) F \left( \ell \right)$.
The probability density function $f(\ell)$ is given by (as you indicated) $ f \left( \ell \right) = \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) $ where the notation $1_{\ell}A$ with interval $A$ is that of an indicator variable. This means $ 1_{\ell} \left( A \right) = \left\{ \begin{array}{lll} 1 & & \text{if } \ell \in A\\ 0 & & \text{otherwise} \end{array} \right. $ Therefore the cumlative distribution function is given by $ F( \ell)= \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) $ Multiplying both we get the density \begin{eqnarray*} 2 f \left( \ell \right) F \left( \ell \right) & = & 2 \left\{ \frac{2}{5} 1_{\ell} \left[ 0, 2 \right) + \left( - \frac{2}{5} \ell + \frac{6}{5} \right) 1_{\ell} \left[ 2, 3 \right) \right\}\\ & \times & \left\{ \frac{2 \ell}{5} 1_{\ell} \left[ 0, 2 \right) + \frac{1}{5} \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) + 1_{\ell} \left[ 3, \infty \right) \right\}\\ & = & \frac{8 \ell}{25} 1_{\ell} \left[ 0, 2 \right) + \frac{4}{25} \left( - \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) 1_{\ell} \left[ 2, 3 \right) \end{eqnarray*} and therefore \begin{eqnarray*} E \left[ \max \left( L_1, L_2 \right) \right] & = & \frac{8}{25} \int_0^2 \ell^2 \mathrm{d} \ell + \frac{4}{25} \int_2^3 \ell \left( - \ell + 3 \right) \left( - \ell^2 + 6 \ell - 4 \right) \mathrm{d} \ell\\ & = & \frac{637}{375}\\ & \approx & 1.69867 \end{eqnarray*}