6.3 Conservative vector fields  (Page 2/16)

We use [link] to calculate ${\displaystyle {\int }_C\text{F}•d\text{r}.}$ Curve C can be parameterized by $\text{r}\left(t\right)=⟨2t,2t⟩,0\le t\le 1.$ Then, $\text{F}\left(\text{r}\left(t\right)\right)=⟨4t,8t⟩$ and ${\text{r}}^{\prime }\left(t\right)=⟨2,2⟩,$ which implies that

Notice that $F=\text{∇}f,$ where $f\left(x,y\right)=x^2+2y^2.$ If we think of the gradient as a derivative, then $f$ is an “antiderivative” of F . In the case of single-variable integrals, the integral of derivative $g^{\prime }\left(x\right)$ is $g\left(b\right)-g\left(a\right),$ where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral F to be $f\left(P_1\right)-f\left(P_0\right),$ where $P_1$ is the endpoint of the curve of integration and $P_0$ is the start point. Notice that this is the case for this example:

and

In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.