Question

If $I=\underset{0}{\overset{1}{\int }}{x}^{x}dx$, then $I$ lies in the interval

• A. $({\left(\frac{1}{e}\right)}^{1/e},1)$
• B. $({\left(\frac{1}{e}\right)}^e,1)$
• C. $(\frac{1}{e},1)$
• D. $(0,1)$

Answer 1

The correct option is A $({\left(\frac{1}{e}\right)}^{1/e},1)$

Given: $I=\underset{0}{\overset{1}{\int }}{x}^{x}dx$
Let $f\left(x\right)=x^x$
$f^{\prime }\left(x\right)=x^x(\ln x+1)\Rightarrow f^{\prime }\left(x\right)=0atx=\frac{1}{e}$
So, $f\left(x\right)$ is minimum at $x=\frac{1}{e}$ and maximum at $x=1$
For $x\in (0,1)$
$m(1-0)<I<M(1-0)\Rightarrow f\left(\frac{1}{e}\right)<I<f\left(1\right)\therefore {\left(\frac{1}{e}\right)}^{1/e}<I<1$