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

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SUMMARY

The integral \(\int_{0}^{\pi} e^{\sin^2 x}\,dx\) is definitively larger than \(\frac{3\pi}{2}\). This conclusion is reached by applying the inequality \(e^x > 1+x\) for \(x = \sin^2 x\), which holds true for all \(x\) except \(x=0\). The evaluation shows that \(\int_{0}^{\pi} e^{\sin^2 x}\,dx\) exceeds the integral \(\int_0^\pi (1 + \sin^2x)\,dx\), which equals \(\frac{3\pi}{2}\). Therefore, the integral of the exponential function with the sine squared argument is greater than the specified constant.

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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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