MCQ
The value of $\int_{\,{e^{ - 1}}}^{\,{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|\,dx} $ is
- A$\frac{3}{2}$
- ✓$\frac{5}{2}$
- C$3$
- D$5$
$ = \int_{{e^{ - 1}}}^1 { - \frac{{\log x}}{x}dx + \int_1^{{e^2}} {\frac{{\log x}}{x}dx} } $
$ = \int_{ - 1}^0 { - zdz + \int_0^2 {zdz} } $,
(Putting ${\log _e}x = z$ ==> $(1/x)\,dx = dz)$
$ = \left[ { - \frac{{{z^2}}}{2}} \right]_{ - 1}^0 + \left[ {\frac{{{z^2}}}{2}} \right]_0^2 = \frac{1}{2} + 2 = \frac{5}{2}$.
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$( S 1): \lim _{ n \rightarrow \infty} \frac{1}{ n ^2}(2+4+6+\ldots \ldots \ldots+2 n)=1$
(S2) : $\lim _{ n \rightarrow \infty} \frac{1}{ n ^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots \ldots .+ n ^{15}\right)=\frac{1}{16}$