Prove, with no knowledge of the decimal value of $\pi$ should be assumed or used that $\displaystyle 1\lt \int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx \lt \frac{2\sqrt{3}}{3}$.

Prove, with no knowledge of the decimal value of $\pi$ should be assumed or used that $\displaystyle 1\lt \int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx \lt \frac{2\sqrt{3}}{3}$.

The solution below is provided by MarkFL:

We are given to prove:

$\displaystyle 1<\int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx<\frac{2}{\sqrt{3}}\tag{1}$

If we move the integral 4 units to the left, and then use the even function rule, and divide through by 2, we obtain:

$\displaystyle \frac{1}{2}<\int_0^1\frac{1}{\sqrt{4-x^2}}\,dx<\frac{1}{\sqrt{3}}$

If we define:

$\displaystyle f(x)=\frac{1}{\sqrt{4-x^2}}$

then there results:

$\displaystyle f'(x)=\frac{x}{(4-x^2)^{\frac{3}{2}}}$

Since the integrand is strictly increasing within the bounds, we know that the integral is greater than the left Riemann sum and less than the right sum. Using 1 partition, we may then write:

$\displaystyle f(0)<\int_0^1\frac{1}{\sqrt{4-x^2}}\,dx$
 
Or:

$\displaystyle \frac{1}{2}<\int_0^1\frac{1}{\sqrt{4-x^2}}\,dx<\frac{1}{\sqrt{3}}$

Shown as desired.