Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1124: 13

$\dfrac{2}{5}(e-1)$

$I=\int_{0}^{1} (t)(t^2)(e^{t^5}) (2t) dt$ $I=2 \int_{0}^{1} t^4 \times (e^{t^5}) (2t) dt$ Plug $a=t^5; da=5t^4 dt$ $=\dfrac{2}{5} \int_{0}^{1} (e^a) da$ $=\dfrac{2}{5}(e^1-e^0)$ $=\dfrac{2}{5}(e-1)$