The value of the integral ∫limits_0^1 e^x^2 d x lies in the interval

Since $e^{x^2}$ is an increasing function on $(0,1)$.

∴  $m=e^0=1$ and $M=e^1=e$, where

$m$ and $M$ are minimum and maximum values of $f(x)=e^{x^2}$ in the interval $(0,1)$ for all $x \in(0,1)$

$\Rightarrow 1(1-0)<\int\limits_0^1 e^{x^2} d x<e(1-0) \Rightarrow 1<\int\limits_0^1 e^{x^2} d x<e$