5.3 The Fundamental Theorem of Calculus/9: Difference between revisions
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<math>\begin{align} | <math>\begin{align} | ||
g(y)= \int_{y}^{2}t^2\sin{(t)}dt \\[2ex] | g(y)&= \int_{y}^{2}t^2\sin{(t)}dt \\[2ex] | ||
&= 1(y^{2}sin{(y)})-0(2^{2}sin{(2)})\\[2ex] | &= 1(y^{2}sin{(y)})-0(2^{2}sin{(2)})\\[2ex] | ||
&= y^{2} sin{(y)} \\[2ex] | &= y^{2} sin{(y)} \\[2ex] | ||
\end(align)</math> | \end(align)</math> | ||
Revision as of 18:49, 25 August 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} g(y)&= \int_{y}^{2}t^2\sin{(t)}dt \\[2ex] &= 1(y^{2}sin{(y)})-0(2^{2}sin{(2)})\\[2ex] &= y^{2} sin{(y)} \\[2ex] \end(align)}