6.5 Average Value of a Function/5: Difference between revisions
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<math> | <math> | ||
f(t) = te^{-t^2},\quad [0, 5], f(t) = \frac{1}{5}\int_{\0}^{\5} | f(t) = te^{-t^2},\quad [0, 5], f(t) = \frac{1}{5}\int_{\0}^{\5}\te^{-t^2}\,dx | ||
</math> | </math> | ||
Revision as of 18:35, 16 December 2022
Failed to parse (syntax error): {\displaystyle f(t) = te^{-t^2},\quad [0, 5], f(t) = \frac{1}{5}\int_{\0}^{\5}\te^{-t^2}\,dx }