# 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.

## Answer and Explanation:

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}

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from ELM: CSU Math Study Guide

Chapter 15 / Lesson 5