If $\lim\limits _{t \rightarrow 0}\left(\int\limits_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}=\frac{\alpha}{5 e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then $\alpha$ is equal to ________________.
Answer (integer)
64
Solution
<p>$1^{\infty}$ form</p>
<p>Now $\mathrm{L}=\mathrm{e}^{\mathrm{t} \rightarrow 0} \frac{1}{\mathrm{t}}\left(\left.\frac{(3 \mathrm{x}+5)^{\mathrm{t}+1}}{3(\mathrm{t}+1)}\right|_0 ^1-1\right)$</p>
<p>$$\begin{aligned}
& =e^{t \rightarrow 0} \frac{8^{t+1}-5^{t+1}-3 t-3}{3 t(t+1)} \\
& =e \frac{8 \ln 8-5 \ln 5-3}{3} \\
& =\left(\frac{8}{5}\right)^{2 / 3}\left(\frac{64}{5}\right)=\frac{\alpha}{5 \mathrm{e}}\left(\frac{8}{5}\right)^{2 / 3}
\end{aligned}$$</p>
<p>On comparing</p>
<p>$\alpha=64$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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