g ( y ) = ∫ 2 y t 2 sin ( t ) d t {\displaystyle g(y)=\int _{2}^{y}t^{2}\sin {(t)}dt}
d d y [ g ( y ) ] = d d y [ ∫ 2 y t 2 sin ( t ) d t ] = 1 ⋅ ( ( y ) 2 sin y ) − 0 ⋅ ( ( 0 ) 2 sin ( 2 ) ) = y 2 sin ( y ) {\displaystyle {\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)}}
Therefore, g ′ ( y ) = y 2 sin ( y ) {\displaystyle {\text{Therefore, }}g'(y)=y^{2}\sin {(y)}}
![{\displaystyle {\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)}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/95fa152b3d3ba4f0fae375aa07a9effd08e165d2)
