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How can I derive the vector equation $$\vec{a}_{avg} = \frac{\vec{v}_2-\vec{v}_1}{t_2-t_1} = \frac{\Delta \vec{v}}{\Delta t}$$ for average acceleration?

I know how to derive the average $x$ acceleration, the average $y$ acceleration and the average $z$ acceleration:

$$a_{x,avg} = {1 \over t_2-t_1} \int_{t_1}^{t_2} a_x(t) dt = {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_x \over dt} dt = {[v_x(t)]_{t_1}^{t_2} \over t_2-t_1} = {v_x(t_2) - v_x(t_1) \over t_2-t_1}$$

and similarly for the average $y$ acceleration $a_{y,avg}$ and the average $z$ acceleration $a_{z,avg}$. However, I don't know how to say that the $x$-component of the average acceleration vector $\vec{a}_{avg}$ equals the average $x$ acceleration, that the $y$-component of the average acceleration vector $\vec{a}_{avg}$ equals the average $y$ acceleration, and that the $z$-component of the average acceleration vector $\vec{a}_{avg}$ equals the average $z$ acceleration. That is, I don't know how to say that:

$$\vec{a}_{avg} = (a_{x,avg},a_{y,avg},a_{z,avg})$$

How can I rigorously show that $\vec{a}_{avg} = (a_{x,avg},a_{y,avg},a_{z,avg})$?

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Since your acceleration vector is $$\vec{a} = \bigg( \frac{d v_x}{dt}, \frac{d v_y}{dt}, \frac{d v_z}{dt} \bigg),$$ averaging over time one gets $$ \vec{a}_{avg} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} \vec{a}(t) dt = \bigg( {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_x \over dt} dt, {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_y \over dt} dt, {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_z \over dt} dt \bigg) = \frac{1}{ t_2-t_1 } \big([v_x(t)]_{t_1}^{t_2}, [v_y(t)]_{t_1}^{t_2}, [v_z(t)]_{t_1}^{t_2} \big) = \frac{\Delta \vec{v}}{\Delta t}.$$

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