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Which of these numbers is larger?

\[\large \int_0^\pi e^{\sin^2 x} \, dx \qquad \text{ OR } \qquad 1.5 \pi \ ? \]

Note by Leo X 2 years, 2 months ago

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Note that \( e^x = 1 + x + \frac{x^2}{2!} + ... > 1 + x \) for \( x > 0 \).

Therefore, \( \int_0^\pi e^{\sin^2 x} dx > \int_0^\pi 1 + \sin^2x dx = \frac{3\pi}{2} \)

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the integral is equal to \(\pi \sqrt { e } { I }_{ 0 }\left( \frac { 1 }{ 2 } \right) \) where \(I_n\) is the modified Bessel function of the first kind,the answer comes out to be \(\approx5.5\) which is greater than \(1.5\pi\)

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestNote that \( e^x = 1 + x + \frac{x^2}{2!} + ... > 1 + x \) for \( x > 0 \).

Therefore, \( \int_0^\pi e^{\sin^2 x} dx > \int_0^\pi 1 + \sin^2x dx = \frac{3\pi}{2} \)

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the integral is equal to \(\pi \sqrt { e } { I }_{ 0 }\left( \frac { 1 }{ 2 } \right) \) where \(I_n\) is the modified Bessel function of the first kind,the answer comes out to be \(\approx5.5\) which is greater than \(1.5\pi\)

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