The smallest interval in which the value of $\int\limits_0^1 \frac{x}{x^3+16} d x$ lies is

$\left[0, \frac{1}{17}\right]$

The function f(x) = $\frac{x}{x^3+16}$ is increasing in [0, 1]

∴ Min. value of f(x) is f(0) = 0 and Max. value of f(x) is f(1) = $\frac{1}{17}$

Since $m(b-a) \leq \int\limits_a^b f(x) d x \leq M(b-a)$,

$0 \leq \int\limits_0^1 \frac{x}{x^3+16} d x \leq \frac{1}{17}$

Hence (1) is the correct answer.