Answer
See proof
Work Step by Step
Step 1: Consider \[ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}, \] \[ \mathbf{r}_1(t) = x_1(t)\mathbf{i} + y_1(t)\mathbf{j}, \] \[ \mathbf{r}_2(t) = x_2(t)\mathbf{i} + y_2(t)\mathbf{j}. \] Step 2 (a): Let \(k\) be a scalar. \[ \begin{aligned} \int_{a}^{b} \mathbf{r}(t) \, dt &= \int_{a}^{b} [x(t) \mathbf{i} + y(t) \mathbf{j}] \, dt \\ &= \int_{a}^{b} x(t) \, dt \mathbf{i} + \int_{a}^{b} y(t) \, dt \mathbf{j} \\ &= k\int_{a}^{b} x(t) \, dt \mathbf{i} + k\int_{a}^{b} y(t) \, dt \mathbf{j} \\ &= k\int_{a}^{b} [x(t) \mathbf{i} + y(t) \mathbf{j}] \, dt \\ &= k\int_{a}^{b} \mathbf{r}(t) \, dt. \end{aligned} \] Step 3 (b): \[ \begin{aligned} \int_{a}^{b} [\mathbf{r}_1(t) + \mathbf{r}_2(t)] \, dt &= \int_{a}^{b} [(x_1(t) + x_2(t)) \mathbf{i} + (y_1(t) + y_2(t)) \mathbf{j}] \, dt \\ &= \int_{a}^{b} [(x_1(t) + x_2(t))] \, dt \mathbf{i} + \int_{a}^{b} [(y_1(t) + y_2(t))] \, dt \mathbf{j} \\ &= \int_{a}^{b} x_1(t) dt \mathbf{i} + y_1(t) dt \mathbf{j} + \int_{a}^{b} x_2(t) \mathbf{i} + y_2(t) \, dt \mathbf{j} \\ &= \int_{a}^{b} \mathbf{r}_1(t) \, dt + \int_{a}^{b} \mathbf{r}_2(t) \, dt. \end{aligned} \] Step 4 (c): \[ \begin{aligned} \int_{a}^{b} [\mathbf{r}_1(t) - \mathbf{r}_2(t)] \, dt &= \int_{a}^{b} [(x_1(t) - x_2(t)) \mathbf{i} + (y_1(t) - y_2(t)) \mathbf{j}] \, dt \\ &= \int_{a}^{b} [(x_1(t) - x_2(t))] \, dt \mathbf{i} + \int_{a}^{b} [(y_1(t) - y_2(t))] \, dt \mathbf{j} \\ &= \int_{a}^{b} [x_1(t) - x_2(t)] \, dt \mathbf{i} + \int_{a}^{b} [y_1(t) - y_2(t)] \, dt \mathbf{j} \\ &= \int_{a}^{b} \mathbf{r}_1(t) \, dt - \int_{a}^{b} \mathbf{r}_2(t) \, dt. \end{aligned} \]
