## WebWork 18.2 #7

This problem is given in differential form. It is stated as follows:

Problem: Evaluate the integral $\int_C 5ydx+7xdy$ where $C$ is the straight line path from $(1,4)$ to $(3,6)$.

Hints: First, you should parameterize the curve. One possibility is $x=1+2t$ and $y=4+2t$ where $0\le t\le 1$. Since we now know what $x$ and $y$ are, the integral can be rewritten

$\int_C 5ydx+7xdy=\int_C 5(4+2t)dx+7(1+2t)dy.$

Now, we should convert this integral to be in terms of $t$ and $dt$, instead of having the $dx$ and $dy$. That’s easy, since $dx=2dt$ and $dy=2dt$ (differentiate $x$ and $y$ with respect to $t$). Now, we’re in this awkward looking situation:

$\int_0^1 5(4+2t)2dt+7(1+2t)2dt$

But, this is just simplified as

$\int_0^1 54+48t dt.$

Now, solve that.