Why do we need an integral to prove that $\frac{22}{7} > \pi$? We know this famous (and beautiful) integral which shows that $\dfrac{22}{7} > \pi$ as : 
$$0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi$$
Now since the integrand is positive, hence:
$$\dfrac{22}{7}-\pi>0$$
$$\color{blue}{\dfrac{22}{7}>\pi}$$

Although I can see its beauty, why is it needed to show that  $\dfrac{22}{7} > \pi$ ?
Can't we just say that : 
$$\dfrac{22}{7}=\color{red}{3.142}857142857142857\cdots = 3. \overline{142857}$$
$$\pi =\color{red}{3.141}592653589793238\cdots$$
And hence it is greater ??

Thanks!!
 A: The point is that $\pi$ is usually defined as the ratio between the circumference of a circle and its diameter, rather than as a particular numerical value. One can compute the numerical value of $\pi$ to some (even arbitrary) precision in many ways, but this particular integral gives a quick way to give an easy but relatively tight upper bound on its value using essentially geometric means, i.e., using the above definition of $\pi$: The appearance of $\pi$ in the value of this integral comes from the occurrence of the arctangent function in the antiderivative of the integrand, and this itself follows quickly from the unit circle definitions of certain trigonometric functions.
Note too that one can quickly a get cheap but good lower bound on $\pi$ by comparing with a similar function:
$$\frac{22}{7} - \pi = \int_0^1 \frac{x^4 (1 - x)^4}{1 + x^2} dx < \int_0^1 x^4 (1 - x)^4 dx = \frac{1}{630} .$$
Rearranging gives that
$$3.14126 \ldots = \frac{1979}{630} < \pi < \frac{22}{7} = 3.14285 \ldots ,$$ and using the midpoint of these bounds gives an estimate for $\pi$ sure to be accurate to within $\frac{1}{1260} < 10^{-3}$.
Applying the same sort of considerations to the integrals $$\int_0^1 \frac{x^{4 n} (1 - x)^{4 n}}{1 + x^2} dx$$ for larger integers $n$ gives even better (and, for $n$ not too large, still relatively efficient) approximations. For example, $n = 2$ yields
$$3.1415917 \ldots = \frac{47171}{15015} < \pi < \frac{3849155}{1225224} = 3.1415928 \ldots ,$$
which yields an estimate accurate within $10^{-6}$.
A: Depending on your priorities, one problem with your explanation is that there are details hiding in how mathematicians found the digits of $\pi$ in the first place. If we make those details explicit, what we get is that we have some series (many different choices will do)
$$\pi = \sum_{i=0}^\infty a_i$$
and a way of bounding our error, so
$$\left|\pi - \sum_{i=0}^n a_i\right| < E_n.$$
If we want to know $\frac{22}{7} > \pi$ explicitly, we can compute $E_n$ and $\sum_{i=0}^n a_i$ for successively larger values of $n$, until we have that $\frac{22}{7}$ exceeds $\sum_{i=0}^n a_i$ by more than the error $E_n$.
By looking up digits of $\pi$ on the computer or on a calculator, you're doing the same thing, looking up previously computed approximations where the value of $n$ is big enough that $E_n$ is within the rounding error for however many digits you asked for.
A: No.  We cannot $\text{“}\,$just say that $\pi=3.141592653589793238\ldots\,{}\text{''}$.  That requires proof!  How can we possibly know that $\pi=3.141592653589793238\ldots$ without doing a lot more work than computing that integral?
A: If you can derive
$\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
then you can extract $\pi<(22/7)$ via a comparison test.  It does, however, take a little patience.
First separate the $n=1$ through $n=4$ terms:
$\dfrac{\pi^2}{6}-\dfrac{205}{144}=\displaystyle \sum_{n=5}^{\infty}\dfrac{1}{n^2}$
Then we have
$\dfrac{1}{n^2}<\dfrac{1}{(n-\frac{1}{2})(n+\frac{1}{2})}=\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}$
$\dfrac{\pi^2}{6}-\dfrac{205}{144}<\displaystyle \sum_{n=5}^{\infty}\left(\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}\right)$
The last sum telescopes to $2/9$.  Therefore
$\dfrac{\pi^2}{6}<\dfrac{205}{144}+\dfrac{2}{9}$
$\pi^2<\dfrac{79}{8}$ 
And then
$\left(\dfrac{22}{7}\right)^2=\dfrac{484}{49}=\dfrac{\color{#0055ff}{3872}}{392}$
$\dfrac{79}{8}=\dfrac{\color{#0055ff}{3871}}{392}$
