Let $$I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$

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

$$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Adding equations $$(1)$$ and $$(2)$$ we get

$$2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$$

$$ = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$$

$$ = 2\int\limits_0^\pi {{{\cos }^2}} x\,dx$$

$$ = 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx$$

$$ = 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$$

$$ \Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx$$

$$ = 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)} $$

$$ \Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx$$

$$ \Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi $$

$$ \Rightarrow I = {\pi \over 2}$$

Given that

$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $$

Differentiating $$w.r.t$$ $$\beta $$

$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $$

$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $$

$$ = 1 - {\pi \over 4} + \sqrt 2 $$

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.