Answer: The exact length of the curve is \(\frac{1}{2}e^2-\frac{1}{6}\).
For a curve with parametric equations \(x(t)\) and \(y(t)\), the arc length from \(t=0\) to \(t=1\) is
\(\displaystyle s=\int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)
Here,
\(x=\frac{1}{2}e^{2t}-\frac{1}{3}t^3-\frac{1}{2}, \qquad y=2e^t(t-1).\)
Differentiating,
\(\displaystyle \frac{dx}{dt}=e^{2t}-t^2\)
and
\(\displaystyle \frac{dy}{dt}=2e^t(t-1)+2e^t=2te^t.\)
Then
\(\begin{aligned} \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 &=(e^{2t}-t^2)^2+4t^2e^{2t}\\ &=e^{4t}-2t^2e^{2t}+t^4+4t^2e^{2t}\\ &=e^{4t}+2t^2e^{2t}+t^4\\ &=(e^{2t}+t^2)^2. \end{aligned}\)
So, for \(0\le t\le 1\),
\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=e^{2t}+t^2.\)
Hence
\(\begin{aligned} s&=\int_0^1 (e^{2t}+t^2)\,dt\\ &=\left[\frac{1}{2}e^{2t}+\frac{1}{3}t^3\right]_0^1\\ &=\left(\frac{1}{2}e^2+\frac{1}{3}\right)-\frac{1}{2}\\ &=\frac{1}{2}e^2-\frac{1}{6}. \end{aligned}\)
Therefore, the exact length is \(\boxed{\frac{1}{2}e^2-\frac{1}{6}}\).
