Answer
$$\frac{15}{32}(e^{16}-e^{64})$$
Work Step by Step
We set up the integral:
$$\int_{C} f(x, y, z) d s=\int_{a}^{b} f(g(t), h(t), k(t))|\mathbf{v}(t)| d t$$
Where
$$f=ye^{x^2}$$ and $$\mathbf{r}(t)=4t\mathbf{i}-3t\mathbf{j}$$ and the $t$ range is $$-1 \leq t \leq 2$$
Thus, we have:
$$\int_{-1}^{2} (-3t)e^{(4t)^2} \sqrt{4^2+(-3)^2} dt = -3 \cdot 5 \int_{-1}^{2} te^{16t^2}dt=\frac{-15}{32}\int_{16}^{64}e^udu=\frac{15}{32}(e^{16}-e^{64})$$
where $u=16t^2$ and $du=32tdt$
