Explanation

(d)
If $f(x)$ is a continuous function defined on $\lbrack a,\ b\rbrack$, then
$m(b - a) \leq \int_{a}^{b}{f(x)dx \leq M(b - a)}$
Where $M$ and $m$ are maximum and minimum values respectively of $f(x)$ in $\lbrack a,b\rbrack$
Here, $f(x) = 1 + e^{- x^{2}}$is continuous on $\lbrack 0,\ 1\rbrack.$
Now, $0 < x < 1\ \Rightarrow x^{2} < x$
$\Rightarrow e^{x^{2}} < e^{x}\ \Rightarrow e^{- x^{2}} > e^{- x}$
Again, $0 < x < 1\ \Rightarrow x^{2} > 0$
$\Rightarrow \ e^{x^{2}} < e^{0}\ \Rightarrow e^{- x^{2}} > 1$
$\therefore e^{- x} < e^{- x^{2}} < 1$ for all $x \in \lbrack 0,\ 1\rbrack$
$\Rightarrow \ 1 + e^{- x} < 1 + e^{- x^{2}} < 2$ for all $x \in \lbrack 0,\ 1\rbrack$
$\Rightarrow \ \int_{0}^{1}{(1 + e^{- x}})dx < \int_{0}^{1}{(1 + e^{- x^{2}}})dx < \int_{0}^{1}{2\ dx}$
$\Rightarrow \ 2 - \frac{1}{e} < \int_{0}^{1}{(1 + e^{- x^{2}}})dx < 2$