5.3 The Fundamental Theorem of Calculus/9

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

Failed to parse (syntax error): {\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)} \\ \text{Therefore, } g'(y) = y^{2}\sin{(y)} }