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\|(\mathrm d\vec r(t))/(\mathrm dt)\right\|=\|3\,\vec\imath-4\,\vec\jmath\|=√(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\|(\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=-(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)}`