Question

The value of the integral ${\int }_0^1{e}^{x^2}dx$ lies in the interval

• A. $(0,1)$
• B. $(-1,0)$
• C. $(1,e)$
• D. None of these

Answer 1

The correct option is C $(1,e)$

Since $e^{x^2}$ is an increasing function on $(0,1)$.
$\therefore m=e^0=1$ and $M=e^1=e$, where
$m$ and $M$ are minimum and maximum values of $f\left(x\right)=e^{x^2}$ in the interval $(0,1)$ for all $x\in (0,1)$

$1<e^{x^2}<e$

Integrating all sides, we get

${\int }_0^{1}1dx<{\int }_0^1{e}^{x^2}dx<{\int }_0^{1}edx$
$\Rightarrow 1(1-0)<{\int }_0^1{e}^{x^2}dx<e(1-0)$

$\Rightarrow 1<{\int }_0^1{e}^{x^2}dx<e$

Hence, option C.