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
$$\eqalign{ & {S_1} = {S_3} = \int\limits_0^4 {y\,dx} = \int\limits_0^4 {\frac{{{x^2}}}{4}dx} = \left[ {\frac{{{x^3}}}{{12}}} \right]_0^4 = \frac{{16}}{3}\,\,{\text{sq}}{\text{.}}\,{\text{units}} \cr & {\text{Also }}{S_2} = \int\limits_0^4 {\left( {2\sqrt x - \frac{{{x^2}}}{4}} \right)dx} = \left[ {\frac{{2{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{x^3}}}{{12}}} \right]_0^4 \cr & = \frac{4}{3} \times 8 - \frac{{16}}{3} = \frac{{16}}{3}\,\,{\text{sq}}{\text{.}}\,{\text{units}} \cr & \therefore {S_1}:{S_2}:{S_3} = 1:1:1 \cr} $$

Given curves $${x^2} = \frac{y}{4}$$  and $${x^2} = 9y$$  are the parabolas whose equations can be written as $$y = 4{x^2}$$  and $$y = \frac{1}{9}{x^2}.$$
Also, given $$y = 2.$$

Now, shaded portion shows the required area which is symmetric.
$$\eqalign{ & \therefore {\text{Area}} = 2\int\limits_0^2 {\left( {\sqrt {9y} - \sqrt {\frac{y}{4}} } \right)dy} \cr & {\text{Area}} = 2\int\limits_0^2 {\left( {3\sqrt y - \sqrt {\frac{y}{2}} } \right)dy} \cr & = 2\left[ {\frac{2}{3} \times 3.{y^{\frac{3}{2}}} - \frac{1}{2} \times \frac{2}{3}.{y^{\frac{3}{2}}}} \right]_0^2 \cr & = 2\left[ {2{y^{\frac{3}{2}}} - \frac{1}{3}{y^{\frac{3}{2}}}} \right]_0^2 \cr & = \left. {2 \times \frac{5}{3}{y^{\frac{3}{2}}}} \right|_0^2 \cr & = 2.\frac{5}{3}.2\sqrt 2 \cr & = \frac{{20\sqrt 2 }}{3} \cr} $$

$$\eqalign{ & {\text{Putting }}x = 0,\,y = 0,\,{\text{we get}}\,{\text{ }}f\left( 0 \right) + f\left( 0 \right) = f\left( 0 \right) \cr & \therefore f\left( 0 \right) = 0 \cr & {\text{Putting }}y = - x,\,f\left( x \right) + f\left( { - x} \right) = f\left( {x - x} \right) = f\left( 0 \right) = 0 \cr & \therefore f\left( { - x} \right) = - f\left( x \right) \cr & \therefore f\left( x \right)\,{\text{is an odd function}}{\text{. So, }}\int_{ - 3}^3 {f\left( x \right)dx = 0} \cr} $$

$$\eqalign{ & {\text{Solving }}y = \sqrt x {\text{ and }}y = \frac{x}{2} - \frac{3}{2},\,\left( {x - 1} \right)\left( {x - 9} \right) = 0 \cr & \therefore x = 9,{\text{ }}1.{\text{ But }}y{\text{ is positive, so }}x \ne 1. \cr & {\text{Hence, }}x = 9. \cr & \therefore {\text{Area}} = \int_0^9 {\sqrt x \,dx - \int_3^9 {\left( {\frac{1}{2}x - \frac{3}{2}} \right)} } dx \cr} $$

The required area is shown by shaded portion in the figure.

The required area is $$A = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\sin \,y} \right|dy = 2\int\limits_0^{\frac{\pi }{2}} {\sin \,y\,dy} = 2} $$

$$\eqalign{ & I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x}}{{{{\cos }^3}x + {{\sin }^3}x}}dx} \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right) + {{\sin }^3}\left( {\frac{\pi }{2} - \pi } \right)}}dx} \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx} \cr & \therefore I + I = \int_0^{\frac{\pi }{2}} {1\,dx} = \frac{\pi }{2} \cr & \therefore I = \frac{\pi }{4} \cr} $$

$$\eqalign{ & I = \int_0^{\frac{\pi }{4}} {\log } \left( {1 + \tan \,x} \right)dx \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log \left\{ {1 + \tan \left( {\frac{\pi }{4} - x} \right)} \right\}dx} \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \left\{ {1 + \frac{{1 - \tan \,x}}{{1 + \tan \,x}}} \right\}dx \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \frac{2}{{1 + \tan \,x}}dx \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \,2\,dx - I \cr & \therefore I = \frac{1}{2}\int_0^{\frac{\pi }{4}} {\log } \,2\,dx = \frac{1}{2}.\log \,2.\frac{\pi }{4} \cr} $$

It is given that, $$f\left( x \right) > x,$$   for all $$x > 1.$$  So, area bounded by $$y = f\left( x \right),\,y = x$$    and the lines $$x = 1,\,x = t$$    is given by $$\int_1^t {\left\{ {f\left( x \right) - x} \right\}dx} $$
But this area is given equal to $$\left( {t + \sqrt {1 + {t^2}} - \sqrt 2 - 1} \right)$$     sq unit.
Therefore, $$\int_1^t {\left\{ {f\left( x \right) - x} \right\}dx = t + \sqrt {1 + {t^2}} - \sqrt 2 - 1,{\text{ for all }}t > 1} $$
On differentiating both sides w.r.t. $$t,$$ we get
$$\eqalign{ & f\left( t \right) - t = 1 + \frac{t}{{\sqrt {1 + {t^2}} }}{\text{ for all }}t > 1 \cr & \Rightarrow f\left( t \right) = t + 1 + \frac{t}{{\sqrt {1 + {t^2}} }}{\text{ for all }}t > 1 \cr & {\text{Hence, }}f\left( x \right) = x + 1 + \frac{x}{{\sqrt {1 + {x^2}} }}{\text{ for all }}x > 1 \cr} $$

$$\eqalign{ & f'\left( x \right) = \log \left( {1 + {x^2}} \right) \cr & \therefore f''\left( x \right) = \frac{{2x}}{{1 + {x^2}}} \cr & \therefore f''\left( 1 \right) = 1 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \int_n^{n + 1} {f\left( x \right)} dx = {n^3}\,;\,n \in {\bf{Z}}......\left( 1 \right) \cr & {\text{Let }}I = \int_{ - 3}^3 {f\left( x \right)dx} \cr & = \int_{ - 3}^{ - 2} {f\left( x \right)} dx + \int_{ - 2}^{ - 1} {f\left( x \right)dx + } \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} + \int_1^2 {f\left( x \right)dx} + \int_2^3 {f\left( x \right)dx} \cr & = {\left( { - 3} \right)^3} + {\left( { - 2} \right)^3} + {\left( { - 1} \right)^3} + {0^3} + {1^3} + {2^3} \cr & = - 27 - 8 - 1 + 0 + 1 + 8 \cr & = - 27 \cr & {\text{Hence, }}\int_{ - 3}^3 {f\left( x \right)dx} = - 27 \cr} $$