Let \( \vec{r}(t)=\left\langle\sqrt[3]{t-4}, 4 e^{3 t}, \frac{2}{t-11}\right\rangle \). Evaluate the definite integral. \( \int_{12}^{31} \vec{r}(t) d t=\langle\square, \square \)

To evaluate the definite integral \( \int_{12}^{31} \vec{r}(t) d t \), we need to integrate each component of the vector function \( \vec{r}(t) \). 1. **First component:** \[ \int_{12}^{31} \sqrt[3]{t-4} \, dt \] Let \( u = t-4 \), then \( du = dt \) and the limits change from \( t = 12 \) to \( u = 8 \) and \( t = 31 \) to \( u = 27 \): \[ \int_{8}^{27} u^{1/3} \, du = \left[ \frac{3}{4} u^{4/3} \right]_{8}^{27} = \frac{3}{4} \left( 27^{4/3} - 8^{4/3} \right) = \frac{3}{4} \left( 81 - 16 \right) = \frac{3}{4} \times 65 = \frac{195}{4} \] 2. **Second component:** \[ \int_{12}^{31} 4 e^{3t} \, dt = 4 \int_{12}^{31} e^{3t} \, dt = \frac{4}{3} e^{3t} \bigg|_{12}^{31} = \frac{4}{3} \left(e^{93} - e^{36}\right) \] 3. **Third component:** \[ \int_{12}^{31} \frac{2}{t-11} \, dt \] This integrates to: \[ 2 \ln |t-11| \bigg|_{12}^{31} = 2 \left( \ln |31 - 11| - \ln |12 - 11| \right) = 2 \left( \ln 20 - \ln 1 \right) = 2 \ln 20 \] Combining all components, we get: \[ \int_{12}^{31} \vec{r}(t) \, dt = \left\langle \frac{195}{4}, \frac{4}{3} \left(e^{93} - e^{36}\right), 2 \ln 20 \right\rangle \] Thus, the final answer is: \[ \int_{12}^{31} \vec{r}(t) d t = \left\langle \frac{195}{4}, \frac{4}{3} \left(e^{93} - e^{36}\right) \right\rangle \]