Question

The value of the definite integral $\int\limits_{0}^{1}\frac{x\,dx}{x^3+16}$ lies in the interval [a, h], The smallest such interval is

• A. $\left[0,\frac{1}{17}\right]$
• B. $[0,1]$
• C. $\left[0,\frac{1}{27}\right]$
• D. none of these,

Answer 1

The Correct Option is A

$f \left(x\right)=\frac{x}{x^{3}+16} \Rightarrow f'\left(x\right)$ $=\frac{\left(x^{3}+16\right)\cdot1-x\cdot3x^{2}}{\left(x^{3}+16\right)^{2}}$ $=\frac{x^{3}-3x^{3}+16}{\left(x^{3}+16\right)^{2}}=\frac{16-2x^{3}}{\left(x^{3}+16\right)^{2}}$ $>\, 0$ for all $x \in\left[0,1\right]$ $\therefore f \left(x\right)$ is monotonic increasing function. Least value $= 0$, greatest value $=\frac{1}{17}$ in $\left[0,1\right]$ Since $m\left(b- a\right) \le \int\limits_{a}^{b} f \left(x\right) \le M \left(b-a\right)$ $\therefore 0\left(1-0\right)\le\int_{0}^{1} \frac{x\,dx}{x^{3}+10}\frac{1}{17} \le\left(1-0\right)$ $\Rightarrow 0 \le\int_{0}^{1}\frac{x\,dx}{x^{3}+16}\le\,\frac{1}{17}$