# How to calculate line integral on polygonal path

Given is vector field $v(x,y)=(y,2x)$. Calculate line integral on polygonal path from $A(1,1)$ to $P(0,1)$ to $Q=(0,4)$ to $B(2,4)$.

What I did: I did parametrization.

1. $A \mapsto P$

$\gamma _1(t)=\begin{pmatrix} 1-t \\ 1 \end{pmatrix}$

2. $P \mapsto Q$

$\gamma _2(t)=\begin{pmatrix} 0 \\ 1+3t \end{pmatrix}$

3. $Q \mapsto B$

$\gamma _3(t)=\begin{pmatrix} 2t \\ 4 \end{pmatrix}$

But what now? I think I should have $3$ integrals and just add them. But I dont know how excatly to get those integrals and what are theirs boundaries?

Should maybe 1. integral be:

$\int_{0}^{1} \! 1 dt + \int_{0}^{1} \! (2-2t)dt \,$

Hint: If $C = C_1 + \cdots C_n$ and each $C_i$ is a piecewise smooth curve then:
$$\int_C f \cdot d\vec{r} = \sum_{i=1}^n \int_{C_i} f \cdot d\vec{r}$$
Recall that if $\gamma_i:[x_{i-1},x_i] \to \mathbb{R}^n$ is a parametrization for $C_i$ then by definition we have:
$$\int_{C_i} f \cdot d\vec{r} = \int_{x_{i-1}}^{x_i} f(\gamma_i) \cdot \gamma'_i(t) \ dt$$
• Sorry, but I dont understand. What is $x_i$ excatly? Commented Dec 11, 2016 at 18:47
• Like this: $\int_{1-t}^{1} \! 1 \cdot 0 \, dt+\int_{1-t}^{1} \! 2(1-t) \,(-2) dt$ ? Commented Dec 11, 2016 at 19:03