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

Revision as of 18:47, 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><br>
-<math>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)})</math><br>
-y^{2} sin{(y)}</math>
+<math>= y^{2} sin{(y)}</math>