Answer
See proof.
Work Step by Step
Step 1: Given \[ \mathbf{F}(x,y) = \cos x \mathbf{i} + \sin x \mathbf{j} \] where \(\mathbf{C}\) is the curve \[ \mathbf{r}(t) = -\frac{\pi }{2}i + tj \quad (1 \leq t \leq 2) \] Step 2: By inspection, the tangent vector is given by \[ \mathbf{T} = \mathbf{j} \] Then, \[ \mathbf{F} \cdot \mathbf{T} = \mathbf{F} \cdot \mathbf{j} = \sin x \] on \(\mathbf{C}\). Since \(x = -\frac{\pi}{2}\) on \(\mathbf{C}\), then \[ \sin x = -1, \quad \mathbf{F} \cdot \mathbf{T} = -1 \] and \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (-1) ds \] Result: See proof.
