$$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $$

$$3{x^2} - 5x + 2 = 0$$

$$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$$

$$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$$

$$ \Rightarrow (x - 1)(3x - 2) = 0$$

$$\therefore$$ In 0 to $${2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2$$

And in $${2 \over 3}$$ to 1, $$|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)$$

$$\therefore$$ $$\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} } $$

$$ = \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} } $$

Now, let $$f(x) = - 3{x^2} + 7x - 1$$

$$ \Rightarrow f'(x) = - 6x + 7$$

$$ = - 6\left( {x - {7 \over 6}} \right)$$

$$f(0) = - 1$$

$$f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1$$

$$ = {7 \over 3} = 2.33$$

$$\therefore$$ Range of $$f(x) = \left[ { - 1,{7 \over 3}} \right]$$

Integer value between $$ - 1$$ to $${7 \over 3} = - 1,0,1,2,{7 \over 3}$$

When $$f(x) = - 1$$

$$ \Rightarrow - 3{x^2} + 7x - 1 = - 1$$

$$ \Rightarrow 3{x^2} - 7x = 0$$

$$ \Rightarrow x(3x - 7) = 0$$

$$ \Rightarrow x = 0,{7 \over 3}$$ (not possible as $${7 \over 3} \notin (0,2)$$)

When $$f(x) = 0$$

$$ \Rightarrow - 3{x^2} + 7x - 1 = 0$$

$$ \Rightarrow 3{x^2} - 7x + 1 = 0$$

$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}$$

$$ = {{7\, \pm \sqrt {37} } \over 6}$$

$$\therefore$$ $$x = {{7 - \sqrt {37} } \over 6}$$

When $$f(x) = 1$$

$$ \Rightarrow - 3{x^2} + 7x - 1 = 1$$

$$ \Rightarrow 3{x^2} - 7x + 2 = 0$$

$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}$$

$$ = {{7\, \pm \,\sqrt {25} } \over 6}$$

$$ = {{7\, \pm \,5} \over 6}$$

$$ \Rightarrow x = 2,\,{1 \over 3}$$

When $$f(x) = 2$$

$$ \Rightarrow - 3{x^2} + 7x - 1 = 2$$

$$ \Rightarrow 3{x^2} - 7x + 3 = 0$$

$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}$$

$$ = {{7\, \pm \,\sqrt {13} } \over 6}$$

$$\therefore$$ $$x = {{7\, - \sqrt {13} } \over 6}$$

$$\therefore$$ Possible values of $$x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}$$

$$\therefore$$ $$\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx} $$

$$ = \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } } $$

$$ = {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]$$

$$ = {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}$$

$$ = {{\sqrt {37} + \sqrt {13} - 8} \over 6}$$

Now, $$\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $$

Let $$f(x) = 3{x^2} - 3x + 3$$

$$f'(x) = 6x - 3 = 3(2x - 1)$$

$$f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}$$

$$f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)$$

$$ = {7 \over 3} = 2.3$$

$$f(1) = 3(1 - 1 + 1)$$

$$ = 3$$

No integer values present in between $${7 \over 3}$$ and 3.

$$\therefore$$ Possible value of $$x = {2 \over 3},\,1$$

So, integration don't break anywhere.

$$\therefore$$ $$\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $$

$$ = \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx} $$

$$ = \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}} $$

$$\therefore$$ Value of Integration

$$ = {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}$$

$$ = {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}$$

$$ = {{\sqrt {37} + \sqrt {13} - 4} \over 6}$$