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

Revision as of 18:52, 25 August 2022

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

@@ Line 1: / Line 1: @@
-<math>\begin{align}
+<math>g(y)= \int_{y}^{2}t^2\sin{(t)}dt
-g(y)&= \int_{y}^{2}t^2\sin{(t)}dt \\[2ex]
+= 1(y^{2}sin{(y)})-0(2^{2}sin{(2)})
-&= 1(y^{2}sin{(y)})-0(2^{2}sin{(2)})\\[2ex]
+= y^{2} sin{(y)}
-&= y^{2} sin{(y)} \\[2ex]
+</math>
-\end(align)</math>