A certain object moves in such a way that its velocity (in m/s) after time t (in s) is given by v (t )equals t squared plus 3 t plus 3.v(t)=t2+3t+3. Find the distance traveled during the first four seconds by evaluating Integral from 0 to 4 (t squared plus 3 t plus 3 )dt.∫04t2+3t+3 dt. Round your answer to the nearest tenth of a meter.