Answer: \(\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{(t-1)e^{-t}}{4(2-t)^3}.\)
The mean value of \(\frac{\mathrm d^2y}{\mathrm dx^2}\) over \(0\leq x\leq\frac74\) is
\(\frac{4e^{-1/2}-3}{21}.\)
Given
\(x=4t-t^2,\qquad y=1-e^{-t},\)
we have
\(\frac{\mathrm dx}{\mathrm dt}=4-2t=2(2-t),\qquad \frac{\mathrm dy}{\mathrm dt}=e^{-t}.\)
Therefore
\(\frac{\mathrm dy}{\mathrm dx}=\frac{e^{-t}}{4-2t}=\frac{e^{-t}}{2(2-t)}.\)
Differentiate with respect to \(x\) by differentiating with respect to \(t\) and dividing by \(\frac{\mathrm dx}{\mathrm dt}\):
\(\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{\frac{\mathrm d}{\mathrm dt}\left(e^{-t}(4-2t)^{-1}\right)}{4-2t}.\)
Now
\(\frac{\mathrm d}{\mathrm dt}\left(\frac{e^{-t}}{4-2t}\right)=\frac{e^{-t}(2t-2)}{(4-2t)^2}.\)
Hence
\(\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{e^{-t}(2t-2)}{(4-2t)^3}=\frac{(t-1)e^{-t}}{4(2-t)^3}.\)
For the mean value with respect to \(x\) over \(0\leq x\leq\frac74\), first find the corresponding \(t\)-limits. Since \(x=4t-t^2\), \(x=0\) gives \(t=0\), and \(x=\frac74\) gives \(t=\frac12\).
The mean value is
\(\frac{1}{7/4}\int_0^{7/4}\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\frac47\int_0^{7/4}\frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx.\)
But
\(\int \frac{\mathrm d^2y}{\mathrm dx^2}\,\mathrm dx=\frac{\mathrm dy}{\mathrm dx}.\)
So
\(\text{mean}=\frac47\left[\frac{\mathrm dy}{\mathrm dx}\right]_{t=0}^{t=1/2}.\)
Now
\(\frac{\mathrm dy}{\mathrm dx}=\frac{e^{-t}}{4-2t}.\)
Therefore
\(\text{mean}=\frac47\left(\frac{e^{-1/2}}{3}-\frac14\right)=\frac{4e^{-1/2}-3}{21}.\)
