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if i m 1 e x m d x where m n then i 10 10 i 9 is e
Question:
If
$I_{m}=\int\limits_{1}^{e}(\ln x)^{m} d x$
, where
$m \in N$
, then
$I_{10}+10 I_{9}$
is equal to -
BITSAT - 2010
BITSAT
Updated On:
Jan 30, 2025
$e^{10}$
$\frac{e^{10}}{10}$
$e$
$e^{ -1}$
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The Correct Option is
C
Solution and Explanation
$I_{10}=\int\limits_{1}^{e} 1 .(\ln x)^{10} d x=\left[(\ln x)^{10} x\right]_{1}^{e}$
$-\int\limits_{-1}^{e} 10(\ln x)^{9} \cdot \frac{1}{x} \cdot x d x$
$=e-0-10 \int\limits_{1}^{e}(\ln x)^{9} d x$
$=e-10 I_{9}+10 I_{9}=e$
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