A certain object moves in such a way that its velocity (in $\mathrm{m} / \mathrm{s}$ ) after time $\mathrm{t}$ (in $\mathrm{s}$ ) is given by $v=t^{2}+2 t+10$. Use integration to find the distance traveled during the first four seconds.
Answer:

Step 1 ：Given that the velocity of an object is represented by the function \(v = t^{2} + 2t + 10\) where \(v\) is the velocity in m/s and \(t\) is the time in seconds.

Step 2 ：The distance traveled by an object is the integral of its velocity with respect to time.

Step 3 ：Therefore, to find the distance traveled during the first four seconds, we need to integrate the velocity function from 0 to 4.

Step 4 ：The integral of \(v = t^{2} + 2t + 10\) from 0 to 4 gives the distance traveled during the first four seconds.

Step 5 ：After calculating, we find that the distance traveled during the first four seconds is \(\frac{232}{3}\) meters.

Step 6 ：Final Answer: The distance traveled during the first four seconds is \(\boxed{\frac{232}{3} \text{ m}}\).