Intersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\left( {0,0} \right)$$ and $$\left( {4,4} \right).$$ The graph is as shown in the figure.



By symmetry, we observe

$${S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} } $$

$$ = \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}$$ sq. unit

Also $${S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}$$

$$\therefore$$ $${S_1}:{S_2}:{S_3} = 1:1:1$$

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



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

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