Expectation of joint life span The life span of a particular mechanical part is a random variable described by the following PDF:

If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.
I can get the PDF:
$$
f_L(l) = 0.4 (0 \leq l \leq 2) \\
f_L(l) = -0.4l + 1.2 (2 < l \leq 3)
$$
and the expectation:
$$
E(l) = \int_0^3 l f_L(l) dl \approx 1.27
$$
I think 'majority' means 2 or more, so we can focus on two parts of the three, and pay no attention to the third. The translation is $E(max(l1, l2))$, how will this be derived I currently have no idea.

Sorry about the misleading remark "$E(max(l_1, l_2))$", it's wrong to neglect the third part, because if that one fails early, then we only need one of the rest to fail.

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*}
