MHB Prove Inequality w/o Knowledge of $\pi$

[anemone]

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

 

[MarkFL]

My solution:

Spoiler

We are given to prove:

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

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

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

If we define:

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

then there results:

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

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:

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

Or:

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

Shown as desired.

 

[Greg]

Nicely done, MarkFL.

 

[anemone]

My solution:

Spoiler

We are given to prove:

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

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

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

If we define:

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

then there results:

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

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:

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

Or:

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

Shown as desired.

Awesome, MarkFL!(Cool) And thanks for participating!

 

[Albert1]

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

my solution:

Spoiler

let $y=x-4 ,$ we have :
$$$\int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx =2\int_{0}^{1}\dfrac{dy}{\sqrt {2^2-y^2}}=2sin^{-1}\dfrac{1}{2}=\dfrac {\pi}{3}---(1)$$$
we know $\pi>3$ and $\sqrt 3>\dfrac{\pi}{2}$
we get $1<(1)=\dfrac {\pi}{3}<\dfrac {2\sqrt 3}{3}$

 

[MarkFL]

my solution:

Spoiler

let $y=x-4 ,$ we have :
$$$\int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx =2\int_{0}^{1}\dfrac{dy}{\sqrt {2^2-y^2}}=2sin^{-1}\dfrac{1}{2}=\dfrac {\pi}{3}---(1)$$$
we know $\pi>3$ and $\sqrt 3>\dfrac{\pi}{2}$
we get $1<(1)=\dfrac {\pi}{3}<\dfrac {2\sqrt 3}{3}$

Spoiler

You have used knowledge of the decimal value of $\pi$...which goes against the given instructions. :)

 

[Albert1]

Spoiler

You have used knowledge of the decimal value of $\pi$...which goes against the given instructions. :)

Spoiler

I know you will question this, the following diagram is the proof of $3<\pi , and \,\,\sqrt 3>\dfrac {\pi}{2}$
no use the knowledge of the decimal value of $\pi$
in fact this diagram can also be used as to prove :$\sqrt 2<\dfrac {\pi}{2}<\sqrt 3$ and $\sqrt 2+\sqrt 3 >\pi$

View attachment 5490

 

[MarkFL]

Spoiler

I know you will question this, the following diagram is the proof of $3<\pi , and \,\,\sqrt 3>\dfrac {\pi}{2}$
no use the knowledge of the decimal value of $\pi$
in fact this diagram can also be used as to prove :$\sqrt 2<\dfrac {\pi}{2}<\sqrt 3$ and $\sqrt 2+\sqrt 3 >\pi$

Spoiler

Nice! Before I decided there must be a way to do this without evaluating the definite integral (anemone picks problems that usually have some kind of neat twist), I did use a hexagon/circle system, but I only established the upper bound using areas...it didn't occur to me to use perimeters for the lower bound. So, I abandoned that method and looked elsewhere. :)

 

[anemone]

Spoiler

I know you will question this, the following diagram is the proof of $3<\pi , and \,\,\sqrt 3>\dfrac {\pi}{2}$
no use the knowledge of the decimal value of $\pi$
in fact this diagram can also be used as to prove :$\sqrt 2<\dfrac {\pi}{2}<\sqrt 3$ and $\sqrt 2+\sqrt 3 >\pi$

my solution:

Spoiler

let $y=x-4 ,$ we have :
$$$\int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx =2\int_{0}^{1}\dfrac{dy}{\sqrt {2^2-y^2}}=2sin^{-1}\dfrac{1}{2}=\dfrac {\pi}{3}---(1)$$$
we know $\pi>3$ and $\sqrt 3>\dfrac{\pi}{2}$
we get $1<(1)=\dfrac {\pi}{3}<\dfrac {2\sqrt 3}{3}$

Thanks Albert for your geometry approach that so effectively avoided in using the value of pi in your solution! Bravo!