Home -> Solved problems -> Prove that pi is less than 22/7

## Prove that $$\pi$$ is less than $$\frac{22}{7}$$

### Solution

To prove that $$\pi<\frac{22}{7}$$, let’s calculate the next integral
$\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=?$
The integral is well defined. Next, let’s use the binomial theorem
$(1-x)^{4}=x^{4}-4 x^{3}+6 x^{2}-4 x+1$
$x^{4}(1-x)^{4}=x^{8}-4 x^{7}+6 x^{6}-4 x^{5}+x^{4}$
Now, let’s use polynomial division, we get
$\frac{x^{4}(1-x)^{4}}{1+x^{2}}=x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}$
$\Rightarrow \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\int_{0}^{1}\left(x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}\right) d x$
$=\frac{x^{7}}{7}-\frac{4 x^{6}}{6}+x^{5}-\frac{4 x^{3}}{3}+4 x-\left.4 \arctan (x)\right|_{0} ^{1}$
$=\frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\left(\frac{\pi}{4}\right)-(0-0+0-0+0-4(0))$
$=\frac{22}{7}-\pi$
$\frac{x^{4}(1-x)^{4}}{1+x^{2}}>0 \quad \forall x \in(0,1)$
$\Rightarrow \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}}>0$
$\Rightarrow \frac{22}{7}-\pi>0$
$\huge \Rightarrow \pi<\frac{22}{7}$
Home -> Solved problems -> Prove that pi is less than 22/7

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Prove that pi is less than 22/7
Home -> Solved problems -> Prove that pi is less than 22/7