5.3 The Fundamental Theorem of Calculus/9: Difference between revisions

Revision as of 16:25, 25 August 2022

$g(y)=\int _{y}^{2}t^{2}\sin {(t)}dt=1(y^{2}sin{y})-0(2^{2}sin{2})=y^{2}sin{y}$

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	<math>g(y)= \int_{y}^{2}t^2\sin{(t)}dt = 1 *y^{2}		<math>g(y)= \int_{y}^{2}t^2\sin{(t)}dt = 1(y^{2}sin{y})-0(2^{2}sin{2})= y^{2} sin{y} </math>
	y^{2} sin{y} </math>