Question

The value of the definite integral ${\int }_0^1\frac{xdx}{x^3+16}$ lies in the interval [a, b]. The smallest such interval is.

• A.
• B. [0, 1]
• C.
  
• D. None of these

Answer 1

The correct option is A

$f\left(x\right)=\frac{x}{x^3+16}\Rightarrow f^{\prime }\left(x\right)=\frac{16-2x^3}{(x^3+16{)}^2}\therefore f^{\prime }\left(x\right)=0\Rightarrow x=2\text{Also}{f}^{\prime \prime }\left(2\right)<0\Rightarrow$ The function $f\left(x\right)=\frac{x}{x^3+16}$ is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = $\frac{1}{17}$
Therefore by the property $m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le M(b-a)$
(where m and M are the smallest and greatest values of function)
$\Rightarrow 0\le {\int }_0^1\frac{x}{x^3+16}dx\le \frac{1}{17}$