Question

Find the maximum and minimum value of ${\int }_0^1\frac{dx}{1+x^p}$, where $PϵR+$

• A. $(\frac{P}{P^2+1},1)$
• B. $(\frac{P}{P+1},1)$
• C. $(\frac{P}{P-1},1)$
• D. $(\frac{P}{P^2-1},1)$

Answer 1

Answer: (B)

$(\frac{P}{P+1},1)$

$f\left(x\right)=\frac{1}{1+x^p}.f^{\prime }\left(x\right)=-\frac{P{x}^{p-1}}{{(1+x^p)}^2}<0$,

$⟹$ Decreasing in $(0,1)$

$⟹m=f\left(1\right)$ and $M=f\left(0\right)$

or $m=\frac{1}{2}$ and $M=1$

$⟹\frac{1}{2}<{\int }_0^1\frac{1}{1+x^p}dx<1$ (i)

But LHS must be $\frac{P}{P+1}$

$\therefore 1-x^{2p}<1⟹(1-x^p)(1+x^p)<1$

$⟹(1-x^p)<\frac{1}{(1+x^p)}$

Integrate both sides from (0,1),
${\int }_0^1(1-x^p)dx<{\int }_0^1\frac{dx}{1+x^p}$

$⟹1-\frac{1}{P+1}<{\int }_0^1\frac{dx}{1+x^p}$

$⟹\frac{P}{P+1}<{\int }_0^1\frac{dx}{1+x^p}$ (ii)

Comparing (i) and (ii), we get
$\frac{P}{P+1}<{\int }_0^1\frac{dx}{1+x^p}<1$