Answer
$\dfrac{2}{5}(e-1)$
Work Step by Step
$I=\int_{0}^{1} (t) \times (t^2) \times (e^{t^5}) \times (2t) dt=2 \int_{0}^{1} t^4 \times (e^{t^5}) (2t) dt$
Suppose $p=t^5$ and $dp=5t^4 dt$
Thus, we have $I=\dfrac{2}{5} \int_{0}^{1} (e^p) dp=\dfrac{2}{5}(e^1-e^0)$
Hence, we have $I=\dfrac{2}{5}(e-1)$
