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

Revision as of 02:49, 26 August 2022

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

@@ Line 1: / Line 1: @@
-<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt
+<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br>
-= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})
+<math>1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})=
-= y^{2} sin{(y)}
+y^{2} sin{(y)}
 </math>