JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 1)
The integral $$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$ loge x dx is equal to :
$$ - {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}$$
$${3 \over 2} - e - {1 \over {2{e^2}}}$$
$${1 \over 2} - e - {1 \over {{e^2}}}$$
$${3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}$$
Explanation
$$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$
Let $${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$$
$$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $$
$$ = {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e$$
Let $${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$$
$$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $$
$$ = {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e$$
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