Question

${\int }_1^4{\left\{x\right\}}^{\left[x\right]}dx$ is equal to (where $[.]$ and $\{.\}$ represent greatest integer function and fractional part function respectively.).

• A. $\frac{13}{12}$
• B. $\frac{1}{12}$
• C. $\frac{5}{12}$
• D. $\frac{17}{12}$

Answer 1

Answer: (A)

$\frac{13}{12}$

We have

$\underset{1}{\overset{4}{\int }}{\left\{x\right\}}^{\left[x\right]}dx$

$=\underset{1}{\overset{2}{\int }}\left(x-1\right)dx+\underset{2}{\overset{3}{\int }}{\left(x-2\right)}^{2}dx+\underset{3}{\overset{4}{\int }}{\left(x-3\right)}^{3}dx$

$Letx-1=t,$

$x-2=u$

$x-3=v$

Now

$=\underset{0}{\overset{2}{\int }}tdx+\underset{0}{\overset{1}{\int }}{u}^{2}dx+\underset{0}{\overset{1}{\int }}{v}^{3}dx$

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$

$=\frac{13}{12}$

Hence,

Option $\left(A\right)$ is the correct answer.