Problem involving mean value theorem for integrals

Question

I am trying to solve the following question: $\def\pih{{\frac{\pi}2}}$

Does there exist $\xi \in [\pih,\pi]$ such that $\int_{\frac{\pi}2}^{\pi}\sqrt{1+x^2}\cos{x}dx = -\sqrt{1+\xi^2}$ ? Prove the result and state all theorems used.

While trying to solve this I've immediately figured out that the mean value theorem for integrals should be applied. Knowing the statement of this theorem it can be proved (this was done as a part of the course) that if $f(x)$ is a continuous function, there exists $c\in [a,b]$ such that $\int_a^bf(x)dx = f(c)\cdot(b-a).$ This would imply that there is $ c\in[\pih,\pi]$ such that $\sqrt{1+c^2}\cos(c)\cdot\pih=-\sqrt{1+\xi^2}$. This is basically as far as I'v come to solving this. I can found bounds for $-\sqrt{1+\xi^2}$ from knowing the bounds for $\xi$. Using Desmos I've figured out which point satisfies $c$ but there is no approach that I can think off that definitely proves the required statement. Could someone be able to help?

hamam_Abdallah · Accepted Answer · 2021-08-27T20:26:40.950

3

hint

Here, you can use the second mean value theorem for integrals, which states that

If $ f $ is continuous at $ [a,b] $ and $ g $ is integrable at $ [a,b] $ satisfying $$(\forall x\in [a,b]) \;\; g(x)\ge 0 $$

THEN

$$\exists \xi\in [a,b]\;\; : \;\; \int_a^bfg=f(\xi)\int_a^bg$$

In your case $$g(x)=-\cos(x)$$

edited Aug 27 '21 at 20:26

answered Aug 27 '21 at 19:55

hamam_Abdallah

62,951

Is there a way to do it without this theorem? It is not a part of the course and I am not sure that it should be used in that case – Đumić Branislav Aug 27 '21 at 20:33
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@ĐumićBranislav Yes, in the equality you wrote, you can find the expression of $\xi$ with respect of $ c $. You square and get $\xi$. – hamam_Abdallah Aug 27 '21 at 20:50
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@hamam_Abdallah Re...welcome :-)...+1. – Sebastiano Aug 27 '21 at 21:31
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@Sebastiano Hi real friend. will Ronaldo go to MU ? – hamam_Abdallah Aug 27 '21 at 21:32
@hamam_Abdallah Very very kind friend; I think to the INTER and not to the MU or Juventus=Rubentus ahahhahah. – Sebastiano Aug 27 '21 at 21:35
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@Sebastiano Best wishes to You and Inter Milano. – hamam_Abdallah Aug 27 '21 at 21:37
@hamam_Abdallah My sincere regards with all my heart. – Sebastiano Aug 27 '21 at 21:37
@hamam_Abdallah But the function is transcendental, I can't find c in close form – Đumić Branislav Aug 27 '21 at 21:48
@hamam_Abdallah but shouldn't I prove that $\xi$ is in the required interval? – Đumić Branislav Aug 27 '21 at 22:02
@ĐumićBranislav Right . You can do it . – hamam_Abdallah Aug 27 '21 at 22:04
@hamam_Abdallah How do I do that if I cannot analytically solve the equality for c? Pardon my ignorance, I really have no clue... – Đumić Branislav Aug 27 '21 at 22:06
Right... got myself confused. Thanks – Đumić Branislav Aug 27 '21 at 22:13

Problem involving mean value theorem for integrals

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