Question

Find the line integral of f(x,y)= ye^x^2 along the curve r(t)= 4ti-3tj, -1<= t <= 1The integral of f is ??

Answer 1

$\displaystyle\int_Cf(x,y)\,\mathrm dS=\int_{-1}^1f(x(t),y(t))\|\vec r'(t)\|\,\mathrm dt$

$=\displaystyle\int_{-1}^1(-3t)e^{(4t)^2}\sqrt{4^2+(-3)^2}\,\mathrm dt$

$=\displaystyle-15\int_{-1}^1 te^{16t^2}\,\mathrm dt$

Let $u=16t^2$, so that $\mathrm du=32t\,\mathrm dt$:

$=\displaystyle-\frac{15}{32}\int_{16}^{16} e^u\,\mathrm du=\boxed0$