5.3 The Fundamental Theorem of Calculus/9

$g(y)=\int _{2}^{y}t^{2}\sin {(t)}dt$

${\frac {d}{dy}}\left[g(y)\right]={\frac {d}{dy}}\left[\int _{2}^{y}t^{2}\sin {(t)}dt\right]=1\cdot ((y)^{2}\sin {y})-0\cdot ((0)^{2}\sin {(2)})=y^{2}\sin {(y)}$

${\text{Therefore, }}g'(y)=y^{2}\sin {(y)}$