Averaging the inductor waveforms
How about this: 



Notice there is a slight variation in the notation of the result.

Edit for clarification and further point:

From the above equations we get:

\begin{equation}
\ &lt;v_{L}(t) &gt; _{T_{s}} = &lt;v_{g}(t) &gt; _{T_{s}} + &lt;v(t) &gt; _{T_{s}}
\end{equation}

This equation implies that the contents are continous for the whole period.
Now everybody agrees on this, right? We further continue the equation and get:

\begin{equation}
\ &lt;v_{g}(t) &gt; _{T_{s}} = d(t) &lt;v_{g}(t) &gt; _{dT_{s}} + d'(t) &lt;v_{g}(t) &gt; _{d'T_{s}}
\end{equation}

\begin{equation}
\ &lt;v(t) &gt; _{T_{s}} = d(t) &lt;v(t) &gt; _{dT_{s}} + d'(t) &lt;v(t) &gt; _{d'T_{s}}
\end{equation}

but (as can be seen on graph) we assumed this:

\begin{equation}
\ v_{g}(t) = 0 \ (d&lt;t&lt;T_{s})
\\
\ v(t) = 0 \ (0&lt;t&lt;d)
\end{equation}

Thus,

\begin{equation}
\ &lt;v_{g}(t) &gt; _{T_{s}} = d(t) &lt;v_{g}(t) &gt; _{dT_{s}} 
\end{equation}

\begin{equation}
\ &lt;v(t) &gt; _{T_{s}} = d'(t) &lt;v(t) &gt; _{d'T_{s}}
\end{equation}

This makes sense because:

\begin{equation}
\ \frac{d(t)}{T_{s}} &lt; 1
\end{equation}

\begin{equation}
\ \frac{d'(t)}{T_{s}} &lt; 1
\end{equation}

The average of those signals taken over a longer time within Ts would drop because they are zero. The maximum average occurs within the zones they are non-zero.

Therefore, what we finally get is:

\begin{equation}
\ &lt;v_{L}(t) &gt; _{T_{s}} = d(t) &lt;v_{g}(t) &gt; _{dT_{s}} + d'(t) &lt;v(t) &gt; _{d'T_{s}}
\end{equation}

Which is not the result found in the slides but I am not sure where I made a mistake in the math... Maybe the slides just have a minor mistake in notation? (It happens)