The integral [_1^e ( x e )^2x - ( e x )^x ,] loge x dx is equal to :

Question

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 :

Accepted Answer

$$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$

Let  $${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$$

$$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $$

$$ = {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e$$

Solutions · 1