STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$\Rightarrow - {e^{ - x}}dx = dt$ रखने पर,

$ - {e^{ - x}}dx = dt$

$I = \int_2^{1 + \frac{1}{e}} {\frac{{(t - 1)( - dt)}}{t}} $

$= \int_2^{1 + \frac{1}{e}} {\left( {\frac{1}{t} - 1} \right)} \,dt$

$ = \left[ {{{\log }_e}t - t} \right]_2^{1 + \frac{1}{e}} $

$= {\log _e}\left( {1 + \frac{1}{e}} \right) - \left( {1 + \frac{1}{e}} \right) - {\log _e}2 + 2$

$ = {\log _e}\left( {\frac{{e + 1}}{{2e}}} \right) - \frac{1}{e} + 1$.