Answer
$\dfrac{\sqrt {14}}{12}(e^6-1)$
Work Step by Step
$I=\int_{0}^{1} (t) \times e^{(2t) \times (3t)} \sqrt{{{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}}dt=\int_{0}^{1} (t) e^{(2t) \cdot (3t)} \sqrt {14} dt$
or, $I=\sqrt {14} \times \int_{0}^{1} (t) e^{6t^2} dt$
Suppose $p=6t^2$ and $ dp=12t dt$
Thus, we have $I=\sqrt {14} \times \int_{0}^{6} \dfrac{e^{p}}{12} dp=\dfrac{\sqrt {14}}{12}[e^a] _{0}^{6}$
Hence, we have $I=\dfrac{\sqrt {14}}{12}(e^6-1)$
