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

Revision as of 02:33, 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)})<math>
= y^{2} sin{(y)}<math>

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