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

The correct option is C $(\frac{P}{P-1},1)$

Given : ${\int }_0^1\frac{dx}{(1+x^p)}p\in R^{+}$

We know

$1-\frac{1}{x^p}<\frac{1}{1+x^p}<1{\int }_0^1(1-\frac{1}{x^p})dx<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<{\int }_0^1\left(dx\right){\int }_0^1\left(dx\right)-{\int }_0^1\left(x^{-p}dx\right)<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<{\int }_0^1\left(dx\right){\left[x\right]}_0^1-{\left[\frac{x^{-p}}{(-p+1)}\right]}_0^1<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<{\left[x\right]}_0^{1}1-0-\frac{1}{(-p+1)}+0<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<[1-0]\frac{-p+1-1}{(-p+1)}<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<1\frac{-p}{(-p+1)}<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<1\frac{p}{(p-1)}<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<1$

Hence the correct answer is $\frac{p}{(p-1)}<{\int }_0^1\left(\frac{dx}{(1+x^p)}\right)<1$