The value of $\int\limits_0^{100}\{\sqrt{x}\} d x$ (where {x} is the fractional part of x) is
Answer & explanation
Correct answer: option 4
Given integral $=\int\limits_0^{100}(\sqrt{x}-[\sqrt{x}]) d x$ (by the def. of {x})
$=\int\limits_0^{100} \sqrt{x} d x-\sum\limits_{i=1}^{10} \int\limits_{(i-1)^2}^{i^2}[\sqrt{x}] d x$
$=\left(\frac{2}{3} x^{3 / 2}\right)_0^{100}-2 \sum\limits_{i=1}^{10} \int\limits_{i=1}^i[t] d t$ where $t^2=x$
$=\frac{2000}{3}-2\left(\int\limits_0^1 0 . d x+\int\limits_1^2 1 . d x+\int\limits_2^3 2 . d x+........+\int\limits_9^{10} 9 . d x\right)$
$=\frac{2000}{3}-2\left([x]_1^2+[2 x]_2^3+[3 x]_3^4+........+[9 x]_9^{10}\right)$
$=\frac{2000}{3}-2\left(\frac{(9)(9+1)}{2}\right)=\frac{1730}{3}$
Hence (4) is the correct answer.