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

Revision as of 02:50, 26 August 2022

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

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 <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>= y^{2} sin{(y)}</math>
+y^{2} sin{(y)}</math>