If [[t]] denotes the greatest integer [ t], then the value of [_0^1[2 x-|3 x^2-5 x+2|+1] d…

Question

If $$[t]$$ denotes the greatest integer $$\leq t$$, then the value of $$\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x$$ is :

Accepted Answer

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

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

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

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

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

$$\therefore$$ 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

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

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

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

Solutions · 1