Suppose we define the definite integral using the following formula (i) $\int\limits_{a}^{b}{f\left( x \right)dx}=\dfrac{b-a}{2}\left( f\left( a \right)+f\left( b \right) \right)$,
(ii) for the accurate result for $c\in \left( a,b \right)$ $F\left( c \right)=\dfrac{c-a}{2}\left( f\left( a \right)+f\left( b \right) \right)+\dfrac{b-c}{2}\left( f\left( b \right)+f\left( c \right) \right)$ and when $c=\dfrac{a+b}{2}$, $\int\limits_{a}^{b}{f\left( x \right)dx}=\dfrac{b-a}{4}\left( f\left( a \right)+f\left( b \right)+2f\left( c \right) \right)$.
Using (i) or (ii) above, find the best upper bound of $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$.
(a) $1+\sqrt{2}$,
(b) $\dfrac{1+\sqrt{2}}{2}$,
(c) $\dfrac{\sqrt{2}-1}{2}$,
(d) $2\left( \sqrt{2}-1 \right)$.

Hint: We start solving the problem by comparing the given problem with the both conditions given in the paragraph. Once we find the suitable condition to calculate the best upper bound of the given integral, we substitute the limits in the condition. After substitution, we make the necessary calculations to get the required value of the best upper bound.

Complete step-by-step answer:
According to the problem, we need to find the best upper bound of $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$ using the condition (i) and (ii) given in the paragraph.
We can see that the condition (ii) can be used only if the value of c is clearly mentioned in the problem, which is not the case here to solve using the condition (ii).
Now, let us use the condition (i) i.e., $\int\limits_{a}^{b}{f\left( x \right)dx}=\dfrac{b-a}{2}\left( f\left( a \right)+f\left( b \right) \right)$ to find the best upper bound for the definite integral $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$.
So, the best upper bound for the integral $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$ is $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}=\dfrac{1-0}{2}\left( f\left( 0 \right)+f\left( 1 \right) \right)$.
$\Rightarrow \int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}=\dfrac{1-0}{2}\left( \sqrt{1+{{0}^{4}}}+\sqrt{1+{{1}^{4}}} \right)$.
$\Rightarrow \int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}=\dfrac{1}{2}\left( \sqrt{1+0}+\sqrt{1+1} \right)$.
$\Rightarrow \int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}=\dfrac{1}{2}\left( \sqrt{1}+\sqrt{2} \right)$.
$\Rightarrow \int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}=\dfrac{\left( 1+\sqrt{2} \right)}{2}$.
So, we have found the best upper bound for the integral $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$ as $\dfrac{\left( 1+\sqrt{2} \right)}{2}$.
∴ The best upper bound for the integral $\int\limits_{0}^{1}{\sqrt{1+{{x}^{4}}}dx}$ is $\dfrac{\left( 1+\sqrt{2} \right)}{2}$.

So, the correct answer is “Option b”.

Note: We can see that we need to follow the conditions given in the paragraph in order to solve the given problem. The value of the best upper bound we just got is an estimate not an absolute value as we are taking the assumptions here. We can also get the upper bound for the given integral by calculating the integral of the limits of the integrand $\sqrt{1+{{x}^{4}}}$ in the given interval $\left( 0,1 \right)$. We use this type of assumption only in the cases where we are unable to reduce the given integrand into the standard forms or simple forms that are feasible for integration.