Find the anti-derivative. \( \int\left(4 t^{3} \vec{\imath}+6 t \vec{\jmath}-4 \sqrt{t} \vec{k}\right) \mathrm{dt}= \) \( t^{[?]} \vec{\imath}+\left[\quad t^{[\quad]} \vec{\jmath}-\frac{[]}{[\square]} t^{3 / 2} \vec{k}+\vec{C}\right. \)

To find the anti-derivative, we'll integrate each component separately. 1. For the first component \( 4t^3 \vec{\imath} \), the integral is: \[ \int 4t^3 \, dt = t^4 \cdot 4 / 4 = t^4 \vec{\imath} \] 2. For the second component \( 6t \vec{\jmath} \), the integral is: \[ \int 6t \, dt = t^2 \cdot 6 / 2 = 3t^2 \vec{\jmath} \] 3. For the third component \( -4\sqrt{t} \vec{k} \), we rewrite \( \sqrt{t} \) as \( t^{1/2} \) and compute the integral: \[ \int -4t^{1/2} \, dt = -4 \cdot \frac{t^{3/2}}{3/2} = -\frac{8}{3}t^{3/2} \vec{k} \] Putting it all together, we have: \[ \int\left(4 t^{3} \vec{\imath}+6 t \vec{\jmath}-4 \sqrt{t} \vec{k}\right) \mathrm{dt} = t^{4} \vec{\imath} + 3t^{2} \vec{\jmath} - \frac{8}{3} t^{3/2} \vec{k} + \vec{C} \] Where: \[ t^{[4]} \vec{\imath} + \left[3 t^{[2]} \vec{\jmath} - \frac{8}{3} t^{3/2} \vec{k} + \vec{C}\right] \]