Required area $$=$$ Area of $$OAB\,+$$   Area of $$ABC$$

Now, Area of $$OAB$$
$$\eqalign{ & = \int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {g\left( x \right)dx} \cr & = \int\limits_0^1 {{x^2}dx} + \int\limits_1^2 {\left( { - x + 2} \right)dx} \cr & = \left. {\frac{{{x^3}}}{3}} \right|_0^1 + \left[ {\frac{{ - {x^2}}}{2} + 2x} \right]_1^2 \cr & = \frac{1}{3} + \left[ {\left( {\frac{{ - 4}}{2} + 4} \right) - \left( {\frac{{ - 1}}{2} + 2} \right)} \right] \cr & = \frac{1}{3} + \frac{1}{2} \cr & = \frac{5}{6}{\text{ sq}}{\text{. unit}} \cr} $$

$$\eqalign{ & I = \int_0^{2\pi } {\frac{{\left( {2\pi - x} \right){{\sin }^{2n}}\left( {2\pi - x} \right)}}{{{{\sin }^{2n}}\left( {2\pi - x} \right) + {{\cos }^{2n}}\left( {2\pi - x} \right)}}dx} \cr & \,\,\,\,\, = 2\pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx - I} \cr & \therefore 2I = 2\pi \int_0^{2\pi } {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\pi .2\int_0^\pi {\frac{{{{\sin }^{2n}}x}}{{{{\sin }^{2n}}x + {{\cos }^{2n}}x}}dx} \cr} $$
( because the integrand is a periodic function of period $$\pi .$$ )
$$\eqalign{ & I = 2\pi \int_{\frac{\pi }{2}}^{ - \frac{\pi }{2}} {\frac{{{{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}{{{{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right) + {{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}} d\left( {\frac{\pi }{2} - z} \right) \cr & \,\,\,\,\, = 2\pi \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}} dz \cr & \,\,\,\,\, = 4\pi \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}dz} \cr & \therefore I = 4\pi I',\,{\text{where}}\,I' = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}z}}{{{{\cos }^{2n}}z + {{\sin }^{2n}}z}}dz} . \cr & \therefore I' = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}{{{{\cos }^{2n}}\left( {\frac{\pi }{2} - z} \right) + {{\sin }^{2n}}\left( {\frac{\pi }{2} - z} \right)}}} dz \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{2n}}z}}{{{{\sin }^{2n}}z + {{\cos }^{2n}}z}}} dz \cr & \therefore I' + I' = \int_0^{\frac{\pi }{2}} {dz} = \frac{\pi }{2}\,\,\,\,\,\therefore I' = \frac{\pi }{4} \cr & \therefore I = 4\pi I'\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow I = {\pi ^2} \cr} $$

$$I = p\int_{ - a}^a {{{\tan }^3}x\,dx} + q\int_{ - a}^a {{{\cos }^2}x\,dx} + r\int_{ - a}^a {\sin \,x\,dx} $$
$$ = p \times 0 + 2q\int_0^a {{{\cos }^2}x\,dx} + r \times 0,$$        because $${\tan ^3}x$$  and $$\sin \,x$$  are odd functions, and $${\cos ^2}x$$  is an even function.

$$\eqalign{ & \int_a^{a + k} {f\left( x \right)} dx = \int_0^{a + k} {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & = \int_0^k {f\left( x \right)} dx + \int_k^{a + k} {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & {\text{For the second integral, put }}x = z + k \cr & {\text{Then }}\int_k^{a + k} {f\left( x \right)} dx = \int_0^a {f\left( {z + k} \right)} dz = \int_0^a {f\left( z \right)} dz \cr & \therefore \int_a^{a + k} {f\left( x \right)} dx = \int_0^k {f\left( x \right)} dx + \int_0^a {f\left( x \right)} dx - \int_0^a {f\left( x \right)} dx \cr & = \int_0^k {f\left( x \right)} dx \cr & {\text{which depends on }}k{\text{ but not on }}a. \cr} $$

The required area is shown by shaded region

$$\eqalign{ & A = \int\limits_1^3 {\left| {x - 2} \right|dx} \,\,\, = 2\int\limits_2^3 {\left( {x - 2} \right)dx} \cr & = 2\left[ {\frac{{{x^2}}}{2} - 2x} \right]_2^3\,\,\, = 1 \cr} $$

$$x + 2{y^2} = 0\,\,\, \Rightarrow {y^2} = - \frac{x}{2}$$
[ Left handed parabola with vertex at $$\left( {0,\,0} \right)$$  ]
$$x + 3{y^2} = 1\,\,\, \Rightarrow {y^2} = - \frac{1}{3}\left( {x - 1} \right)$$
[Left handed parabola with vertex at $$\left( {1,\,0} \right)$$  ]
Solving the two equations we get the points of intersection as $$\left( { - 2,\,1} \right),\left( { - 2 - 1} \right)$$

The required area is $$ACBDA,$$   given by
$$\eqalign{ & = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right| \cr & = \left| {\left[ {y - \frac{{5{y^3}}}{3}} \right]_{ - 1}^1} \right| \cr & = \left| {\left( {1 - \frac{5}{3}} \right) - \left( { - 1 + \frac{5}{3}} \right)} \right| \cr & = 2 \times \frac{2}{3} = \frac{4}{3}\,{\text{sq}}{\text{. units}} \cr} $$

$$\eqalign{ & I = \int_{ - 2}^0 {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}dx} + \int_0^2 {\frac{{{{\sin }^2}x}}{{\left[ {\frac{x}{\pi }} \right] + \frac{1}{2}}}dx} \cr & \,\,\,\,\, = \int_{ - 2}^0 {\frac{{{{\sin }^2}x}}{{ - 1 + \frac{1}{2}}}dx} + \int_0^2 {\frac{{{{\sin }^2}x}}{{0 + \frac{1}{2}}}dx} \cr & \,\,\,\,\, = 2\int_0^{ - 2} {{{\sin }^2}x\,dx} + 2\int_0^2 {{{\sin }^2}x\,dx} \cr & \,\,\,\,\, = 2\int_0^2 {{{\sin }^2}\left( { - x} \right)\,d\left( { - x} \right)} + 2\int_0^2 {{{\sin }^2}x\,dx} \cr & \,\,\,\,\, = 0 \cr} $$

$$\eqalign{ & y = {\left( {\sqrt a - \sqrt x } \right)^2} \cr & = a + x - 2\sqrt {ax} \cr & {\text{Hence the required area}} \cr & = \int_0^a {y.dx} \cr & = \int_0^a {a + x - 2\sqrt {ax} .dx} \cr & = \left[ {ax + \frac{{{x^2}}}{2} - \frac{{4x\sqrt {ax} }}{3}} \right]_0^a \cr & = {a^2} + \frac{{{a^2}}}{2} - \frac{{4{a^2}}}{3} \cr & = \frac{{3{a^2}}}{2} - \frac{{4{a^2}}}{3} \cr & = \frac{{9{a^2} - 8{a^2}}}{6} \cr & = \frac{{{a^2}}}{6}\,\,{\text{sq}}{\text{.}}\,{\text{units}} \cr} $$

$$\eqalign{ & {\text{In }}\left( {0,\,1} \right),\,{e^{ - {x^2}}} > {e^{ - {x^2}}}{\cos ^2}x > 0\,\,\,\,\,\,\,\,\therefore {I_1} > {I_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{e^{ - {x^2}}} > {e^{ - x}} > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore {e^{ - {x^2}}}{\cos ^2}x > {e^{ - x}}{\cos ^2}x > 0 \cr & \therefore {I_3} > {I_2} \cr & \therefore {I_2} < {I_3} < {I_1} \cr} $$

Required area is equal to the area under the curves $$y \geqslant {x^2}$$  and $$yd''x + 2$$
$$\eqalign{ & \therefore {\text{Required area :}} \cr & {\text{ = }}\int\limits_{ - 1}^2 {\left( {\left( {x + 2} \right) - {x^2}} \right)dx} \cr & {\text{ = }}\left( {\frac{{{x^2}}}{2} + 2x - \frac{{{x^3}}}{3}} \right)_{ - 1}^2 \cr & {\text{ = }}\left( {2 + 4 - \frac{8}{3}} \right) - \left( { + \frac{1}{2} - 2 + \frac{1}{3}} \right) \cr & {\text{ = }}\frac{9}{2} \cr} $$