Question

The value of ${\int }_1^4{\left\{x\right\}}^{\left[x\right]}dx$ (where $[.]$ and $\{.\}$ denotes the greatest integer and fractional part of $x$) is equal to

• A. $\frac{11}{12}$
• B. $\frac{13}{12}$
• C. $\frac{7}{12}$
• D. $\frac{19}{12}$

Answer 1

The correct option is B $\frac{13}{12}$

${\int }_1^4{\left\{x\right\}}^{\left[x\right]}dx={\int }_1^2\left({x-\left[x\right]}^{\left[x\right]}\right)dx+{\int }_2^3\left({x-\left[x\right]}^{\left[x\right]}\right)dx+{\int }_3^4\left({x-\left[x\right]}^{\left[x\right]}\right)dx={\int }_1^2(x-1)dx+{\int }_2^3\left({[x-2]}^2\right)dx+{\int }_3^4\left({[x-3]}^3\right)dx$
$={\left(\frac{{(x-1)}^2}{2}\right)}_1^2+{\left(\frac{{(x-2)}^3}{3}\right)}_2^3+{\left(\frac{{(x-3)}^3}{4}\right)}_3^4$
$=(\frac{1}{2}-0)+(\frac{1}{3}-0)+(\frac{1}{4}-0)=\frac{13}{12}$