# Consider the following line integral, where C is the line segment from (2, 4) to (1, 5)....

## Question:

Consider the following line integral, where {eq}C {/eq} is the line segment from {eq}(2,\ 4) {/eq} to {eq}(1,\ 5) {/eq}.

{eq}\displaystyle \int_C (9 x - 6 y)\ dx + (10 y - 6 x)\ dy {/eq}.

(a) Show that the line integral is path independent by finding a scalar potential function.

(b) Evaluate the integral along the oriented curve by any appropriate method.

## Line Integral:

We will find the potential function by integrating the partial derivatives and then adding the integration constant and to solve the problem we will use the line integral.

Let us find the potential function:

{eq}f_{x}=9x-6y\\ f_{y}=10y-6x\\ f=\int (9x-6y)dx=\frac{9x^{2}}{2}-6xy+c_{x}\\ f=\int (10y-6x)dy=5y^{2}-6xy+c_{y}\\ f=\frac{9x^{2}}{2}-6xy+5y^{2}+c {/eq}

b) Now let us solve the line integral:

{eq}\frac{x-2}{-1}=\frac{y-4}{1}=t\\ x=-t+2\\ y=t+4\\ dx=-dt\\ dy=dt {/eq}

The integral will be:

{eq}=\int_{0}^{1}\left (9(-t+2)-6(t+4) \right )(-dt)+\left ( 10(t+4)-6(-2t) \right )dt\\ =\int_{0}^{1}(31t+34)dt {/eq}

Integrating we get:

{eq}=\frac{31t^{2}}{2}+34t {/eq}

Now let us plug-in the bounds:

{eq}=\frac{99}{2} {/eq}