Find the anti-derivative. $$ \int(2 t-1) \vec{\imath}+4 t^{3} \vec{\jmath}+3 \sqrt{t} \vec{k} d t= $$ $$ \left(t^{[?]}-[\quad]\right) \vec{\imath}+t^{[]]}+[\quad] t^{[] / 2} \vec{k}+\vec{C} $$

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To find the anti-derivative of the vector function

$$
\int(2 t-1) \vec{\imath}+4 t^{3} \vec{\jmath}+3 \sqrt{t} \vec{k} \, dt,
$$

we will integrate each component separately.

1. **Integrate the $$\vec{\imath}$$ component:**
   $$
   \int (2t - 1) \, dt = t^2 - t + C_1.
   $$

2. **Integrate the $$\vec{\jmath}$$ component:**
   $$
   \int 4t^3 \, dt = t^4 + C_2.
   $$

3. **Integrate the $$\vec{k}$$ component:**
   $$
   \int 3\sqrt{t} \, dt = \int 3t^{1/2} \, dt = 3 \cdot \frac{t^{3/2}}{3} = t^{3/2} + C_3.
   $$

Now, we can combine these results into the final expression for the anti-derivative:

$$
\int(2 t-1) \vec{\imath}+4 t^{3} \vec{\jmath}+3 \sqrt{t} \vec{k} \, dt = \left(t^{2} - t\right) \vec{\imath} + t^{4} \vec{\jmath} + t^{3/2} \vec{k} + \vec{C}.
$$

Now, we can fill in the blanks in the expression:

$$
\left(t^{[2]} - [t]\right) \vec{\imath} + t^{[4]} + [1] t^{[3/2]} \vec{k} + \vec{C}.
$$

Thus, the final answer is:

$$
\int(2 t-1) \vec{\imath}+4 t^{3} \vec{\jmath}+3 \sqrt{t} \vec{k} \, dt = \left(t^{2} - t\right) \vec{\imath} + t^{4} \vec{\jmath} + t^{3/2} \vec{k} + \vec{C}.
$$
The anti-derivative is $$ \left(t^{2} - t\right) \vec{\imath} + t^{4} \vec{\jmath} + t^{3/2} \vec{k} + \vec{C} $$.

Find the anti-derivative. \( \int(2 t-1) \vec{\imath}+4 t^{3} \vec{\jmath}+3 \sqrt{t} \vec{k} d t= \) \( \left(t^{[?]}-[\quad]\right) \vec{\imath}+t^{[]]}+[\quad] t^{[] / 2} \vec{k}+\vec{C} \)

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