\( \int_1^4\displaystyle\frac{\{x\}}{[x]}\,dx \) is equal to
[Where \( \{x\} \) = fractional part function and \( [x] \) = greatest integer function]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{11}{12} \)
On each interval [n, n + 1), \( [x]=n \) and \( \{x\}=x-n \). So the integral splits at 2 and 3:
\( \displaystyle \int_1^4\frac{\{x\}}{[x]}\,dx=\int_1^2\frac{x-1}{1}\,dx+\int_2^3\frac{x-2}{2}\,dx+\int_3^4\frac{x-3}{3}\,dx \)
Each \( \int_n^{n+1}(x-n)\,dx=\frac{1}{2} \), so
\( \displaystyle =\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}\right) \)
\( \displaystyle =\frac{1}{2}\times\frac{11}{6} \)
\( \displaystyle =\frac{11}{12} \)