Answer is: 64
\(\frac{\alpha}{5 e}=\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\int_0^1(3 x+5)^t d x-1\right)\right) \)
\(=\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{(3 x+5)^{t+1}}{3(t+1)}\right)_0^1-1\right) \)
\(=\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}}{3(t+1)}-1\right)\right) \)
\(=\exp \left(\lim _{t \rightarrow 0} \frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}-3 t-3}{3(t+1)}\right)\right) \)
\( =\exp \left(\lim _{t \rightarrow 0}\left(\frac{8^{t+1} \cdot \ell n 8-5^{t+1} \ell n 5-3}{3(t+1)}\right)\right) \)
\(=\exp \left(\frac{\ell n 8^8-\ell n 5^5-3}{5}\right) \)
\( =\left(\frac{8}{5}\right)^{2 / 3} \frac{\alpha}{5 e}=\exp \left(\frac{\ell n \left(\frac{8^8}{5^5}\right)}{5}-1\right) \)
\( \Rightarrow\left(\frac{8}{5}\right)^{2 / 3} \frac{\alpha}{5}=\left(\frac{8^8}{5^5}\right)^{1 / 3}=\left(\frac{8^6 \cdot 8^2}{5^3 \cdot 5^2}\right)^{1 / 3}=\frac{64}{5}\left(\frac{8}{5}\right)^{2 / 3} \)
\(\Rightarrow \alpha=64\)
