Given curves are $$y = \sqrt x .....(1)$$
and $$\,2y - x + 3 = 0.....(2)$$
On solving both we get $$y = - 1,\,3$$

Required area :
$$\eqalign{ & = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}dy} \cr & = \left. {{y^2} + 3y - \frac{{{y^3}}}{3}} \right|_0^3 \cr & = 9 \cr} $$

Curve tracing, $$y = x\,{\log _e}x$$
Clearly $$x > 0,$$
For $$0 < x < 1,\,x\,{\log _e}x < 0,{\text{ and for }}x > 1,\,x\,{\log _e}x > 0$$
Also $$x\,{\log _e}x = 0{\text{ or }}x = 1$$
Further $$\frac{{dy}}{{dx}} = 0 \Rightarrow 1 + {\log _e}x = 0{\text{ or }}x = \frac{1}{e},$$       which is point of minima.

Required area
$$\eqalign{ & = \int\limits_0^1 {\left( {2x - 2{x^2}} \right)dx} - \int\limits_0^1 {x\,\log \,x\,dx} \cr & = \left[ {{x^2} - \frac{{2{x^3}}}{3}} \right]_0^1 - \left[ {\frac{{{x^2}}}{2}\log \,x - \frac{{{x^2}}}{4}} \right]_0^1 \cr & = \left( {1 - \frac{2}{3}} \right) - \left[ {0 - \frac{1}{4} - \frac{1}{2}\mathop {\lim }\limits_{x \to 0} {x^2}\log \,x} \right] \cr & = \frac{1}{3} + \frac{1}{4} \cr & = \frac{7}{{12}} \cr} $$

$$\eqalign{ & \int_1^3 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} = \int_1^2 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} + \int_2^3 {\left| {2 - x} \right|\,\left| {{{\log }_e}x} \right|dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_1^2 {\left( {2 - x} \right){{\log }_e}x\,dx} + \int_2^3 { - \left( {2 - x} \right){{\log }_e}x\,dx} \cr & {\text{Now, }}\int {\left( {2 - x} \right)} {\log _e}x\,dx = {\log _e}x.\left( {2x - \frac{{{x^2}}}{2}} \right) - \int {\left( {2x - \frac{{{x^2}}}{2}} \right).\frac{1}{x}dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2x - \frac{{{x^2}}}{2}} \right){\log _e}x - \int {\left( {2 - \frac{x}{2}} \right)dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2x - \frac{{{x^2}}}{2}} \right){\log _e}x - \left( {2x - \frac{{{x^2}}}{4}} \right) \cr & \therefore I = \left[ {\left( {2x - \frac{{{x^2}}}{2}} \right){{\log }_e}x - \left( {2x - \frac{{{x^2}}}{4}} \right)} \right]_1^2 + \left[ { - \left( {2x - \frac{{{x^2}}}{2}} \right){{\log }_e}x + \left( {2x - \frac{{{x^2}}}{4}} \right)} \right]_2^3 \cr & = \left( {4 - 2} \right){\log _e}2 - \left( {4 - 1} \right) + \left( {2 - \frac{1}{4}} \right) + \left\{ { - \left( {6 - \frac{9}{2}} \right){{\log }_e}3 + \left( {6 - \frac{9}{4}} \right) + \left( {4 - 2} \right){{\log }_e}2 - \left( {4 - 1} \right)} \right\} \cr & = 2{\log _e}2 - 3 + \frac{7}{4} - \frac{3}{2}{\log _e}3 + \frac{{15}}{4} + 2{\log _e}2 - 3 \cr & = 4{\log _e}2 - \frac{3}{2}{\log _e}3 - \frac{1}{2} \cr & = \log \frac{{16}}{{3\sqrt 3 }} - \frac{1}{2} \cr} $$

$$y = \frac{{36}}{{{x^2}}},\,x = 6,\,x = 9$$
Desired area $$ = \int\limits_6^9 {\frac{{36}}{{{x^2}}}dx} = - 36\left[ {\frac{1}{9} - \frac{1}{6}} \right] = 2$$

$$\eqalign{ & f\left( a \right) = \int\limits_a^{2a} {\left( {\frac{x}{6} + \frac{1}{{{x^2}}}} \right)dx} \cr & = \left( {\frac{{{x^2}}}{{12}} - \frac{1}{x}} \right)_a^{2a} \cr & = \left( {\frac{{4{a^2}}}{{12}} - \frac{1}{{2a}} - \frac{{{a^2}}}{{12}} + \frac{1}{a}} \right) \cr & = \frac{{{a^2}}}{4} + \frac{1}{{2a}} \cr & {\text{Let }}f'\left( a \right) = \frac{{2a}}{4} - \frac{1}{{2{a^2}}} = 0 \cr & \Rightarrow a = 1{\text{ which is a point of minima}}{\text{.}} \cr} $$

$$\eqalign{ & I = \left[ x \right]_0^1 + \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx} = 1 + \int_0^1 {\frac{1}{{{e^{{x^2}}}}}} dx \cr & {\text{In }}\left( {0,\,1} \right),\,\frac{1}{e} < \frac{1}{{{e^{{x^2}}}}} < 1 \cr & \therefore \int_0^1 {\frac{1}{e}dx < \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx < } \int_0^1 {1\,dx} } \cr & \therefore \frac{1}{e} < \int_0^1 {\frac{1}{{{e^{{x^2}}}}}dx} < 1 \cr & \therefore 1 + \frac{1}{e} < \int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)dx} < 1 + 1 \cr} $$

The given parabola is $${\left( {y - 2} \right)^2} = x - 1$$
Vertex $$\left( {1,\,2} \right)$$  and it meets $$x$$-axis at $$\left( {5,\,0} \right)$$
Also it gives $${y^2} - 4y - x + 5 = 0$$
So, that equation of tangent to the parabola at $$\left( {2,\,3} \right)$$  is
$$\eqalign{ & y.3 - 2\left( {y + 3} \right) - \frac{1}{2}\left( {x + 2} \right) + 5 = 0 \cr & {\text{or }}x - 2y + 4 = 0 \cr} $$
which meets $$x$$-axis at $$\left( { - 4,\,0} \right).$$
In the figure shaded area is the required area.
Let us draw PD perpendicular to $$y$$-axis.

$$\eqalign{ & {\text{Then required area :}} \cr & = {\text{ Ar }}\Delta {\text{BOA}} + {\text{Ar }}\left( {{\text{OCPD}}} \right) - {\text{Ar }}\left( {\Delta {\text{APD}}} \right) \cr & = \frac{1}{2} \times 4 \times 2 + \int_0^3 {x\,dy} - \frac{1}{2} \times 2 \times 1 \cr & = 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy} \cr & = 3 + \left[ {\frac{{{{\left( {y - 2} \right)}^3}}}{3} + y} \right]_0^3 \cr & = 3 + \left[ {\frac{1}{3} + 3 + \frac{8}{3}} \right] \cr & = 3 + 6 \cr & = 9{\text{ sq}}{\text{. units }} \cr} $$

Required area
$$\eqalign{ & = \int\limits_1^e {\ln \,y\,dy} \cr & = \left( {y\,\ln \,y - y} \right)_1^e \cr & = \left( {e - e} \right) - \left[ { - 1} \right] \cr & = 1 \cr} $$

Also, $$\int\limits_1^e {\ln \,y\,dy} = \int\limits_1^e {\ln \left( {e + 1 - y} \right)dy} $$
Further, required area $$ = e \times 1 - \int\limits_0^1 {{e^x}dx} $$

Given that $$\int\limits_{\frac{\pi }{4}}^\beta {f\left( x \right)dx} = \beta \,\sin \,\beta + \frac{\pi }{4}\cos \,\beta + \sqrt 2 \beta $$
Differentiating w.r.t $$\beta $$
$$\eqalign{ & f\left( \beta \right) = \beta \,\cos \,\beta + \sin \,\beta - \frac{\pi }{4}\sin \,\beta + \sqrt 2 \cr & f\left( {\frac{\pi }{2}} \right) = \left( {1 - \frac{\pi }{4}} \right)\sin \frac{\pi }{2} + \sqrt 2 = 1 - \frac{\pi }{4} + \sqrt 2 \cr} $$

$$x{y^2} = {a^2}\left( {a - x} \right) \Rightarrow x = \frac{{{a^3}}}{{{y^2} + {a^2}}}$$
The given curve is symmetrical about $$x$$-axis, and meets it at $$\left( {a,\,0} \right).$$

The line $$x = 0,$$  i.e., $$y$$-axis is an asymptote
$$\therefore $$  Area is
$$\eqalign{ & = \int\limits_0^\infty {x\,dy} \cr & = 2\int\limits_0^\infty {\frac{{{a^3}}}{{{y^2} + {a^2}}}dx} \cr & = 2{a^2}\frac{1}{a}\left[ {{{\tan }^{ - 1}}\frac{y}{a}} \right]_0^\infty \cr & = 2{a^2}\frac{\pi }{2} \cr & = \pi {a^2}{\text{ sq}}{\text{. units}} \cr} $$