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Let $I_{n}=\int_{0}^{1}(\log x)^{n} d x$, where $n$ is a non-negative integer. Then $I_{2001}-2011 I_{2010}$ is equal to-
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Verified Answer
The correct answer is:
$\mathrm{I}_{100}+100 \mathrm{I}_{99}$
$\quad I_{n}=\int_{1}^{e} 1 .\left(\log _{1} x\right)^{n} d x$
$$
\begin{array}{l}
I_{n}=\left.(\log x)^{n} x\right|_{1} ^{e}-\int_{1}^{e} \frac{n(\log x)^{n-1}}{x} \cdot x d x \\
I_{n}=e-0-n I_{n-1} \\
I_{n}+n I_{n-1}=e \\
I_{2001+2011} \quad I_{2010}=e \\
\square_{100}+100 \mathrm{I}_{99}=e
\end{array}
$$
$$
\begin{array}{l}
I_{n}=\left.(\log x)^{n} x\right|_{1} ^{e}-\int_{1}^{e} \frac{n(\log x)^{n-1}}{x} \cdot x d x \\
I_{n}=e-0-n I_{n-1} \\
I_{n}+n I_{n-1}=e \\
I_{2001+2011} \quad I_{2010}=e \\
\square_{100}+100 \mathrm{I}_{99}=e
\end{array}
$$
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