Proving an Inequality: A Scientific Approach

[AlexOliya]

Homework Statement

Homework Equations

With the regards to posting such a incomplete equation, I will soon put in the updated one
Thank you

The Attempt at a Solution

visual graph... didn't help

 

[pasmith]

What is the range of integration?

 

[AlexOliya]

What is the range of integration?

Our teacher did not write any range :/

 

[Ray Vickson]

Homework Statement

Homework Equations

The Attempt at a Solution

visual graph... didn't help

PF Rules require you to make some efforts and to show your work before seeking help (or, more precisely, they do not permit us to help until you have shown your work).

Anyway, the question is incomplete: what are the integration limits?

Note added in edit: I see that the latter question has already been posed to you, but appeared on my screen only after I pressed the 'enter' key.

 

[AlexOliya]

PF Rules require you to make some efforts and to show your work before seeking help (or, more precisely, they do not permit us to help until you have shown your work).

Anyway, the question is incomplete: what are the integration limits?

Certainly I didn't tackle the equation without effort but I didn't want to send my rather nonsensical and messy attempts .
I replied the last question you asked above

 

[Ray Vickson]

Certainly I didn't tackle the equation without effort but I didn't want to send my rather nonsensical and messy attempts .
I replied the last question you asked above

If you do not show your "nonsensical and messy" attempts, how are we supposed to know where you are having trouble? We need to see what you have done, wrong or not! (BTW: I edited my previous post, which dealt with your response about integration limits).

 

[AlexOliya]

If you do not show your "nonsensical and messy" attempts, how are we supposed to know where you are having trouble? We need to see what you have done, wrong or not! (BTW: I edited my previous post, which dealt with your response about integration limits).

Okay haha!
Well I looked for the first function which turned out a complex function. and I graphed the function using the Algeo app but still didn't help. I tried setting the limits pi fourth to half pi but still is risky
Our teacher suggested a guide tip for using the Mean value theorem... But it doesn't have a range !

 

[micromass]

How can you even begin to solve this if you don't know the integration limits? This is crucial information

 

[Ray Vickson]

Okay haha!
Well I looked for the first function which turned out a complex function. and I graphed the function using the Algeo app but still didn't help. I tried setting the limits pi fourth to half pi but still is risky
Our teacher suggested a guide tip for using the Mean value theorem... But it doesn't have a range !

Those limits cannot possibly work, as they would involve values of $x$ for which $\sin^{-1} (x)$ is either undefined, or not real. (I am assuming you mean the inverse sine function, rather than the reciprocal $1/\sin x$.)

 

[AlexOliya]

How can you even begin to solve this if you don't know the integration limits? This is crucial information

rather should I storm upon the teacher for this crucial mistake?

 

[micromass]

Yes.

 

[AlexOliya]

Those limits cannot possibly work, as they would involve values of $x$ for which $\sin^{-1} (x)$ is either undefined, or not real. (I am assuming you mean the inverse sine function, rather than the reciprocal $1/\sin x$.)

definitely I mean the inverse sine function. Well it has come to a conclusion that limits are needed and essential

 

[AlexOliya]

Yes.

Thanks for all the help

 

[Ray Vickson]

definitely I mean the inverse sine function. Well it has come to a conclusion that limits are needed and essential

If $f(x) = \arcsin(x)/(1+x^8)$, the possible case $\int_{-1}^1 f(x) \, dx$ is easy to compute explicitly (with almost no work), and the left-hand inequality is FALSE but the right-hand one is (trivially) true. Thus, if the integration limits are -1 and +1, only one of the two inequalities is true.

The other obvious possibility, $\int_0^1 f(x) \, dx$, takes more work; both bounds are true, as shown by numerical evaluation of the integral. However, proving it "analytically" is a challenge.

 

[AlexOliya]

If $f(x) = \arcsin(x)/(1+x^8)$, the possible case $\int_{-1}^1 f(x) \, dx$ is easy to compute explicity (with almost no work), and the inequality is (trivially) true in that case. The other obvious possibility, $\int_0^1 f(x) \, dx$, takes more work; the lower bound is easy to verify (again, with essentially no work), while the upper bound is true (as seen numerically), but is not at all obvious anymore. Why don't you see if you can prove $\int_0^1 f(x) \, dx \leq \frac{\pi}{2} - 1$, without performing a numerical evaluation of the integral?

I did try this today in fact. Tomorrow I will see my Math teacher and definitely will ask why she made this mistake and see what she had in mind!
Thank you for your help dear friend

 

[Ray Vickson]

I did try this today in fact. Tomorrow I will see my Math teacher and definitely will ask why she made this mistake and see what she had in mind!
Thank you for your help dear friend

Please note that I made an error in post #14 (due to a typo). I have edited that post, and you should look at it again.

 

[AlexOliya]

Please note that I made an error in post #14 (due to a typo). I have edited that post, and you should look at it again.

I come updated.
The limits are 0 to positive 1

 

[AlexOliya]

Homework Statement

Homework Equations

The integration limit is 0 to +1

The Attempt at a Solution

visual graph... didn't help

 

[fresh_42]

It's pretty easy if you apply the mean value theorem for integrals (and I made no mistake).
Maybe some work has to be done to show the upper bound is proper.

https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration

 

[SammyS]

For the upper bound, compare to $\displaystyle \ \int_0^1 {\sin^{-1\,}\!(x)}\ dx \$.

 

[Ray Vickson]

For the upper bound, compare to $\displaystyle \ \int_0^1 {\sin^{-1\,}\!(x)}\ dx \$.

Works for the lower bound as well.

 

[AlexOliya]

It's pretty easy if you apply the mean value theorem for integrals (and I made no mistake).
Maybe some work has to be done to show the upper bound is proper.

https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration

Thank you!

 

[AlexOliya]

Works for the lower bound as well.

Many Many thanks! Helped a lot!