Question

The smallest interval $[a,b]$ such that $\int\limits_{0}^{1}{\dfrac{dx}{\sqrt{1+{{x}^{4}}}}}\in [a,b]$ is given by
A. $\left[ \dfrac{1}{\sqrt{2}},1 \right]$
B. $[0,1]$
C. $\left[ \dfrac{1}{2},2 \right]$
D. $\left[ \dfrac{3}{4},1 \right]$

Answer 1

Hint: In this question, we are to find the smallest interval that satisfies the given integral. For this, the limits of the given integral are altered in such a way that we get the required smallest interval for the given integral.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ - lower limit and $b$- upper limit.
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$



Complete step by step solution:Consider the given integral as
$I=\int\limits_{0}^{1}{\dfrac{dx}{\sqrt{1+{{x}^{4}}}}}$
The given integral has the interval as $0\le x\le 1$
For the required integral, we can alter the interval as
$0\le x\le 1$
On squaring, we get
$\Rightarrow 0\le {{x}^{2}}\le 1$
Again, on squaring, we get
$\Rightarrow 0\le {{x}^{4}}\le 1$
Adding $1$ on both sides,
$\begin{align}
  & \Rightarrow 1+0\le 1+{{x}^{4}}\le 1+1 \\
 & \Rightarrow 1\le 1+{{x}^{4}}\le 2 \\
\end{align}$
Applying square root, we get
$\begin{align}
  & \Rightarrow \sqrt{1}\le \sqrt{1+{{x}^{4}}}\le \sqrt{2} \\
 & \Rightarrow 1\le \sqrt{1+{{x}^{4}}}\le \sqrt{2} \\
\end{align}$
We need the required variable in the denominator. So, if the terms in inequality are inverted or reciprocated then the inequality gets changed. I.e.,
\[\begin{align}
  & \Rightarrow 1\le \sqrt{1+{{x}^{4}}}\le \sqrt{2} \\
 & \Rightarrow \dfrac{1}{1}\ge \dfrac{1}{\sqrt{1+{{x}^{4}}}}\ge \dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \dfrac{1}{\sqrt{2}}\le \dfrac{1}{\sqrt{1+{{x}^{4}}}}\le 1 \\
\end{align}\]
Thus, we can write,
$\begin{align}
  & \Rightarrow \dfrac{1}{\sqrt{2}}\le \int\limits_{0}^{1}{\dfrac{dx}{\sqrt{1+{{x}^{4}}}}}\le 1 \\
 & \therefore I\in \left[ \dfrac{1}{\sqrt{2}},1 \right] \\
\end{align}$
Here the smallest interval is defined for an integral as
$m(b-a)\le \int\limits_{a}^{b}{f(x)dx\le M(b-a)}$
Where $m(b-a)$ is the smallest value of $f(x)$ and $M(b-a)$ is the highest value of $f(x)$.



Option ‘A’ is correct



Note: Here we need to remember that the required limits are obtained by altering the limits of the given integral.