First we draw each curve as separate graph



NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.

Clearly the bounded area is as shown in the following figure.



Required area $$ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$$

$$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units

We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$

$$\int\limits_0^2 {xf'\left( x \right)} dx$$

$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$

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

$$ = 2f\left( 2 \right) - {3 \over 4}$$

$$ = 0 - {3 \over 4}$$

$$\left( {} \right.$$ As $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$ = - {3 \over 4}.$$

$$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$$

$$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$$

$$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$$

$$\left[ \, \right.$$ as $$\int\limits_{ - a}^a {f\left( x \right)} dx = 0$$ $$\left. \, \right]$$

if $$f(x)$$ is odd

$$ = 2\int\limits_0^a {f\left( x \right)} dx$$ if $$f(x)$$ is even.

$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$$

$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} $$

$$ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} $$

$$ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$$

put $$\cos x = t \Rightarrow - \sin xdx = dt$$

$$\therefore$$ $$I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt$$

$$ = 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt$$

$$ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$$

$$ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]$$

$$ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}$$

$$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$$

$$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $$

$$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$$

$$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$

$$ = 5 - \sqrt 3 - \sqrt 2 $$