Which is Larger: \(\int_{0}^{\pi} e^{\sin^2 x}\,dx\) or \(\frac{3\pi}{2}\)?

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Discussion Overview

The discussion centers on comparing the value of the integral \(\int_{0}^{\pi} e^{\sin^2 x}\,dx\) with \(\frac{3\pi}{2}\). Participants explore inequalities and mathematical reasoning related to this comparison.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose using the inequality \(e^x > 1+x\) for \(x = \sin^2 x\) to argue that \(\int_{0}^{\pi} e^{\sin^2 x}\,dx\) is greater than \(\int_0^\pi (1 + \sin^2x)\,dx\).
  • Participants calculate \(\int_0^\pi (1 + \sin^2x)\,dx\) to show it equals \(\frac{3\pi}{2}\), suggesting that \(\int_{0}^{\pi} e^{\sin^2 x}\,dx > \frac{3\pi}{2}\).
  • There are multiple instances of similar reasoning presented, with slight variations in the phrasing of the inequality and the conclusion drawn.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using the inequality to compare the two values, but the discussion does not reach a consensus on the actual comparison of the integral and \(\frac{3\pi}{2}\).

Contextual Notes

The discussion relies on the validity of the inequality \(e^x > 1+x\) and the calculations of the integral \(\int_0^\pi (1 + \sin^2x)\,dx\), which may depend on specific assumptions about the behavior of the functions involved.

anemone
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Which of the following is the bigger?

$$\int_{0}^{\pi} e^{\sin^2 x}\,dx$$ and $\dfrac{3\pi}{2}$
 
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anemone said:
Which of the following is the bigger?

$$\int_{0}^{\pi} e^{\sin^2 x}\,dx$$ and $\dfrac{3\pi}{2}$
[sp]
Replace $x$ by $\sin^2x$ in the inequality $e^x > 1+x$ (which holds for all $x$ except $x=0$) to see that $$\int_{0}^{\pi} e^{\sin^2 x}\,dx > \int_0^\pi (1 + \sin^2x)\,dx = \pi + \frac\pi2 = \frac{3\pi}2.$$[/sp]
 
Last edited:
Opalg said:
[sp]
Replace $x$ by $\sin^2x$ in the inequality $e^x > 1+x$ (which holds for all $x$ except $x=0$) to see that $$\int_{0}^{\pi} e^{\sin^2 x}\,dx > \int_0^\pi (1 + \sin^2x)\,dx = 1 + \frac\pi2 = \frac{3\pi}2.$$[/sp]

Small typo, it should read $\pi + \frac{\pi}{2}$ :)
 
Opalg said:
[sp]
Replace $x$ by $\sin^2x$ in the inequality $e^x > 1+x$ (which holds for all $x$ except $x=0$) to see that $$\int_{0}^{\pi} e^{\sin^2 x}\,dx > \int_0^\pi (1 + \sin^2x)\,dx = \pi + \frac\pi2 = \frac{3\pi}2.$$[/sp]

Well done, Opalg!(Happy) And thanks for participating!;)
 

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