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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
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
    $5$
✓

Answer

Correct option: B.
$\frac{5}{2}$
b
(b) $\int_{{e^{ - 1}}}^{{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|dx = \int_{{e^{ - 1}}}^1 {\left| {\frac{{{{\log }_e}x}}{x}} \right|\,dx + \int_1^{{e^2}} {\left| {\frac{{{{\log }_e}x}}{x}} \right|\,dx} } } $

$ = \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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