$\int\limits_0^4\{\sqrt{x}\}$ is equal to, where x denotes the fraction part of x.
Answer & explanation
Correct answer: option 4
We have,
$\int\limits_0^4\{\sqrt{x}\} d x=\int\limits_0^1 \sqrt{x} d x+\int\limits_1^4(\sqrt{x}-1) d x=\frac{2}{3}+\frac{2}{3}(8-1)-3=\frac{7}{3}$