The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$ is as shown in the fig.



Required area

$$A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx$$

put $$x + e = t \Rightarrow dx = dt$$

also At $$x = 1 - e,t = 1$$

At $$x = 0,\,\,t = e$$

$$\therefore$$ $$A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]$$

$$e - e - 0 + 1 = 1$$

Hence the required area is $$1$$ square unit.

$${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$$

$$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$

$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$

$$\forall 0 < x < 1,\,{x^2} > {x^3}$$

$$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$$

and $$\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $$

$$ \Rightarrow {I_1} > {I_2}$$ and $${I_4} > {I_3}$$

The required area is shown by shaded region



$$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$$

$$ = 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$$

$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$

$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$