Solving the RMS of waveform
Note that in your case:

$$\int_{0}^{1}v(t-2)^{2}\:\mathrm{d}t=\int_{2}^{3}v(t)^{2}\:\mathrm{d}t$$

Using this, and the generic formula for a parabola with it's vertex at the origin, we have that: \$v(t-2)=4t^{2}\$, we therefore have that:

$$\int_{2}^{3}(v(t))^{2}\:\mathrm{d}t=\int_{0}^{1}(4t^{2})^{2}\:\mathrm{d}t=\int_{0}^{1}16t^{4}\:\mathrm{d}t=\left[\frac{16t^{5}}{5}\right]_{0}^{1}=\frac{16}{5}$$

I hope this helps you!



EDIT: In order to clarify how I arrived at the equation \$v(t-2)=4t^{2}\$, we can examine the general equation of a parabola, given by: $$y(x)=a(x-b)^{2}+c$$

Where \$a\$ is a parameter which modifies the gradient of the parabola and \$b\$ and \$c\$ determine the vertex (bottom-/top-most point) of the parabola (at \$(x,y)=(b,-c)\$), in this case we have that the vertex of \$v(t-2)\$ is at the point \$(0,0)\$ (imagine shifting the graph 2 units to the left) and therefore \$b=-c=0\$ and therefore we have the equation: $$v(t-2)=a(t-2)^{2}$$

Using the fact that when \$t-2=1\$, \$v(t-2)=4\$, we get \$a=4\$.