A ball is spun with angular acceleration α = 6 t 2 - 2 t where t is in second and α is in rad…

Given: $\alpha =6t^2-2t$

Using relation $\alpha =\frac{d\omega }{dt}=6t^2-2t$

Integrating the above, 

${\int }_{10}^{\omega }d\omega ={\int }_0^t\left(6t^2-2t\right)dt$

$\Rightarrow \omega -10=2t^3-t^2$

Now, $\omega =\frac{d\theta }{dt}=10+2t^3-t^2$

${\int }_4^{\theta }d\theta ={\int }_0^t\left(10+2t^3-t^2\right)dt$

Integrating the above relation, 

$\theta -4=10t+\frac{t^4}{2}-\frac{t^3}{3}$

Thus, angular position of the ball is $\theta =\frac{t^4}{2}-\frac{t^3}{3}+10t+4$.