Answer
$\int_C \mathbf{F} \cdot d\mathbf{r} = -140$
Work Step by Step
Step 1: Note that the line integral with respect to \(\vec{x}\) and \(\vec{y}\) of the vector field \(\mathbf{F}(x,y) = f(x,y)\mathbf{i} + g(x,y)\mathbf{j}\) along \(\mathbf{C}\) is expressed as: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (f(x,y)\mathbf{i} + g(x,y)\mathbf{j}) \cdot (d\mathbf{xi} + d\mathbf{yj}) = \int_C f(x,y)dx + g(x,y)dy \] Where, \(\vec{r} = x\mathbf{i} + y\mathbf{j}\) Step 2: Now, let's use \((1)\) in order to evaluate the given integral. Hence, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (-8x\mathbf{i} + 3y\mathbf{j}) \cdot (d\mathbf{xi} + d\mathbf{yj}) \] Therefore \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (-8x\mathbf{i} + 3y\mathbf{j}) \cdot (d\mathbf{xi} + d\mathbf{yj}) \] Step 3: Note that this integral is along the line segment \(\mathbf{C}\) from \(P(-1,0)\) to \(Q(6,0)\). Thus, \(\Delta x = x_Q - x_P = 6 - (-1) = 7\) \(\Delta y = y_Q - y_P = 0 - 0 = 0\) Therefore, \(\Delta \mathbf{r} = \Delta x \mathbf{i} + \Delta y \mathbf{j} = 7\mathbf{i} + 0\mathbf{j} = 7\mathbf{i}\) Then, let's substitute \(\Delta x = 7\) and \(\Delta y = 0\) in \((2)\). Hence, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (-8 \cdot 7\mathbf{i} + 3 \cdot 0\mathbf{j}) \cdot (7\mathbf{i}) \] Step 4: We notice that this integral is only with respect to \(\mathbf{x}\), then its limits are \(x_P = -1\) and \(x_Q = 6\). Thus, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{-1}^{6} (-8 \cdot 7\mathbf{i}) \cdot (7\mathbf{i})\,d\mathbf{x} \] Result: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = -140 \]
