$$I = \int\limits_a^b {xf\left( x \right)} dx$$

$$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$$

$$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$$

$$ = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx} $$

$$\left[ {} \right.$$ As given that $$f\left( {a + b - x} \right) = f\left( x \right)$$ $$\left. {} \right]$$

$$2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx$$

$$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$$

Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$

Integrating log

$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$

$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$

$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$

$$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$

$$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$

$$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$

$$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$

$$ = e - {{{e^2}} \over 2} - {3 \over 2}$$

$$A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } $$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } $$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} $$

$$ = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } } $$

$$ = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2$$

$$ = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)$$

$$ = 1 + 2 + 4 - 3 = 4$$ sq. units

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