If I_1 = _e^ e^2 dx x and I_2 = _1^2 e^x x ,dx, then - Vidyadip

MCQ

If ${I_1} = \int_e^{{e^2}} {\frac{{dx}}{{\log x}}} $ and ${I_2} = \int_1^2 {\frac{{{e^x}}}{x}\,dx,} $ then

✓
${I_1} = {I_2}$
B
${I_1} > {I_2}$
C
${I_1} < {I_2}$
D
None of these

✓

Answer

Correct option: A.

${I_1} = {I_2}$

a
(a) Put $\log x = u$ in ${I_1},$

so that $dx = x\,du = {e^u}du$

Also as $x = e$ to ${e^2},u = 1$ to $2$

Thus, ${I_1} = \int_1^2 {\frac{{{e^u}}}{u}du = \int_1^2 {\frac{{{e^x}}}{x}dx} } $.

Hence, ${I_1} = {I_2}$.

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