Question

The value of ${\int }_0^2\left[x^2-x+1\right]dx$, where $[.]$ denotes the greatest integer function) is given by

• A. $\frac{5-\sqrt{5}}{2}$
• B. $\frac{6-\sqrt{5}}{2}$
• C. $\frac{7-\sqrt{5}}{2}$
• D. $\frac{8-\sqrt{5}}{2}$

Answer 1

The correct option is A $\frac{5-\sqrt{5}}{2}$

$f\left(x\right)=x^2-x+1$

clearly at x = 0, f(0) =1

and at x = 1, f(0) = 1

also $f^{\prime }\left(x\right)=2x-1=0$

at $x=\frac{1}{2}$

hence min value of f(x) at x = 1/2

and is equal to $f\left(1/2\right)=3/4$

now

$I={\int }_0^2[x^2-x+1]dx={\int }_0^2\left[\right(x-1/2{)}^2+3/4]dx$

consider cases for

$3/4⩽(x-\frac{1}{2}{)}^2+3/4⩽1$ at $xϵ[0,1]$

$1⩽(x-\frac{1}{2}{)}^2+3/4⩽2$ at $xϵ[1,\frac{\sqrt{5}+1}{2}]$

$2⩽(x-\frac{1}{2}{)}^2+3/4<3$ at $xϵ[\frac{\sqrt{P}5+1}{2},2]$

$\therefore I={\int }_0^{1}0dx+{\int }_1^{\frac{\sqrt{5}+1}{2}}1dx+{\int }_{\frac{\sqrt{5}+1}{2}}^{2}2dx$

$=\frac{\sqrt{5}+1}{2}-1+2[2-(\frac{\sqrt{5}+1}{2}){]}^2$

$=-\frac{\sqrt{5}}{2}+\frac{5}{2}=\frac{5-\sqrt{5}}{2}$