The value of the definite integral $\int_0^1\frac{x\,dx}{(x^3+16)}$ lies in the interval [a,b]. Then smallest such interval is

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

We have

$∵(1 – 0)(\frac{0}{0^3+16})≤\int_0^1\frac{x\,dx}{(x^3+16)}$

$≤(1 – 0)(\frac{1}{1^3+16})$  (by property)

$0≤\int_0^1\frac{x\,dx}{x^3+16}≤\frac{1}{17}∴\int_0^1\frac{x\,dx}{x^3+16}∈[0,\frac{1}{17}]$