Question

Find the line integral of f(x, y) = y e^X^2 along the curve r(t) = 3t i - 4t J, -2 lessthanorequalto t lessthanorequalto -1. The integral of f is (Type an exact answer.)

Answer 1

$\vec r(t)=3t\,\vec\imath-4t\,\vec\jmath$

$\implies\left\|\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\right\|=\|3\,\vec\imath-4\,\vec\jmath\|=\sqrt{3^2+(-4)^2}=5$

The line integral of $f(x,y)$ along the curve $C$ parameterized by $\vec r(t)$ is

$\displaystyle\int_Cf(x,y)\,\mathrm ds=\int_{-2}^{-1}f(3t,-4t)\left\|\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\right\|\,\mathrm dt=-20\int_{-2}^{-1}te^{9t^2}\,\mathrm dt$

Substitute $u=9t^2$ so that $\mathrm du=18t\,\mathrm dt$. Then

$\displaystyle\int_Cf(x,y)\,\mathrm dS=-\frac{20}{18}\int_{36}^9e^u\,\mathrm du=\frac{10}9\int_9^{36}e^u\,\mathrm du=\boxed{\frac{10}9(e^{36}-e^9)}$