How do I compute the average and RMS of a series of triangular pulses?
Average current is the sum of the current over a cycle, divided by the length of the cycle

Your current is a triangle with \$I_{peak}=1.8\$ amps, and a base of about \$t_{on}=.9\$ divisions, so you get an area of \$\frac{1}{2}BH=.81\$ (some unit of charge, since it's current x time). You then spread that out over \$T=3.7\$ divisions (charge/time = current again), which gives an average of .22 amps.

Alternately, if you work the calculus to get a general form for the average current of a series of triangular pulses:

\$I(t)=I_{peak}(1-\frac{t}{t_{on}})\$ between \$t=0\$ and \$t=t_{on}\$, and zero elsewhere. 

\$I_{AVG}=\frac{\int_{0}^{T}{I(t) dt}}{T}\$ by definition of mean.

Substituting in I(t) from above:

\$I_{AVG}=\frac{\int_{0}^{T}{I_{peak}(1-\frac{t}{t_{on}}) dt}}{T}\$.

Rearranging and executing integral:

\$I_{AVG}=\frac{I_{peak}}{T}(t-\frac{t^2}{2t_{on}})]^{t_{on}}_{0}\$

\$I_{AVG} = \frac{I_{peak}D}{2}\$

Where \$D=\frac{t_{on}}{T}\$. This gives us the same answer as above, for your duty cycle of about .25. 

RMS current is a little more complicated.

\$I_{RMS}=\sqrt{\frac{\int_{0}^{T}{I(t)^2 dt}}{T}}\$ by definition of RMS.

Substituting in I(t) from above:

\$I_{RMS}=\sqrt{\frac{\int_{0}^{t_{on}}{I_{peak}^2(1-\frac{t}{t_{on}})^2 dt}}{T}}\$ 

Rearranging:

\$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{\int_{0}^{t_{on}}{(1-\frac{t}{t_{on}})^2 dt}}\$ 

\$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{
\int_{0}^{t_{on}}{(1-\frac{2t}{t_{on}} + \frac{t^2}{t_{on}^2}) dt}
}\$ 

\$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{
(t-\frac{t^2}{t_{on}} + \frac{t^3}{3t_{on}^2})]^{t_{on}}_{0}
}\$ 

\$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{
\frac{t_{on}}{3}
}\$ 

\$I_{RMS} = \frac{I_{peak}\sqrt{D}}{\sqrt{3}}\$

So your RMS is about .52 amps.