Evaluate the integral \( \int_{0}^{1}\left[\left(4 t^{2}\right) i+(3) j+(3 t+5) k\right] d t \)

To evaluate the integral \[ \int_{0}^{1}\left[\left(4 t^{2}\right) i+(3) j+(3 t+5) k\right] d t, \] we can break it down into its component parts. The integrand is a vector function, and we can integrate each component separately. The vector function can be expressed as: \[ \mathbf{F}(t) = (4t^2) \mathbf{i} + 3 \mathbf{j} + (3t + 5) \mathbf{k}. \] We will integrate each component from \( t = 0 \) to \( t = 1 \). 1. **Integrating the \( i \) component:** \[ \int_{0}^{1} 4t^2 \, dt. \] To compute this integral, we find the antiderivative: \[ \int 4t^2 \, dt = \frac{4}{3} t^3 + C. \] Now we evaluate it from 0 to 1: \[ \left[ \frac{4}{3} t^3 \right]_{0}^{1} = \frac{4}{3}(1^3) - \frac{4}{3}(0^3) = \frac{4}{3} - 0 = \frac{4}{3}. \] 2. **Integrating the \( j \) component:** \[ \int_{0}^{1} 3 \, dt. \] The antiderivative is: \[ \int 3 \, dt = 3t + C. \] Evaluating from 0 to 1 gives: \[ \left[ 3t \right]_{0}^{1} = 3(1) - 3(0) = 3 - 0 = 3. \] 3. **Integrating the \( k \) component:** \[ \int_{0}^{1} (3t + 5) \, dt. \] The antiderivative is: \[ \int (3t + 5) \, dt = \frac{3}{2} t^2 + 5t + C. \] Evaluating from 0 to 1 gives: \[ \left[ \frac{3}{2} t^2 + 5t \right]_{0}^{1} = \left( \frac{3}{2}(1^2) + 5(1) \right) - \left( \frac{3}{2}(0^2) + 5(0) \right) = \left( \frac{3}{2} + 5 \right) - 0 = \frac{3}{2} + 5 = \frac{3}{2} + \frac{10}{2} = \frac{13}{2}. \] Now, we can combine the results of the integrals for each component: \[ \int_{0}^{1} \mathbf{F}(t) \, dt = \left( \frac{4}{3} \right) \mathbf{i} + 3 \mathbf{j} + \left( \frac{13}{2} \right) \mathbf{k}. \] Thus, the final result of the integral is: \[ \int_{0}^{1}\left[\left(4 t^{2}\right) i+(3) j+(3 t+5) k\right] d t = \frac{4}{3} \mathbf{i} + 3 \mathbf{j} + \frac{13}{2} \mathbf{k}. \]