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

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<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>

Revision as of 18:52, 25 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(y)=\int _{y}^{2}t^{2}\sin {(t)}dt=1(y^{2}sin{(y)})-0(2^{2}sin{(2)})=y^{2}sin{(y)}}

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