JEE MAIN - Mathematics Hindi (2019 - 12th January Evening Slot - No. 1)
समाकलन $$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$ loge x dx का मान है :
$$ - {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}$$
$${3 \over 2} - e - {1 \over {2{e^2}}}$$
$${1 \over 2} - e - {1 \over {{e^2}}}$$
$${3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}$$
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