6.5 Average Value of a Function/5

Failed to parse (syntax error): {\displaystyle f(t) = te^{-t^2} \quad [0, 5] \\ \begin{align} f_{avg} = \frac{1}{5}\int_{0}^{5}te^{-t^2}\,dx = \frac{1}{5}\int_{0}^{5}-\frac{1}{2}(e^u)\,du = \frac{1}{5}\int_{0}^{25}e^u(-\frac{1}{2}du) = \frac{1}{10}\int_{-25}^{0}e^u\,du = \frac{1}{10}e^u \bigg|_{-25}^{0} = \frac{1}{10}-\frac{1}{10}e^{-25} = \frac{1}{10}(1-e^{-25}) \end{align} }