Let \(\displaystyle f\left( x \right) =\int _{ 0 }^{ x }{ t\sin { \frac { 1 }{ t } }\ dt } \), then the number of points of discontinuity of \(f\left( x \right)\) in \(\left( 0,\pi \right) \) is \(\text{__________} \).

We have \(f'(x)=x \sin \frac{1}{x}\), which exists and is continuous everywhere in \((0,\pi)\) (Check that \(\lim_{x \to 0}x \sin \frac{1}{x}=0\)). Thus the function \(f(x)\) has no point of discontinuity in the interval \((0,\pi)\).

Easy Math Editor

`*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

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

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

Sort by:

TopNewestWe have \(f'(x)=x \sin \frac{1}{x}\), which exists and is continuous everywhere in \((0,\pi)\) (Check that \(\lim_{x \to 0}x \sin \frac{1}{x}=0\)). Thus the function \(f(x)\) has no point of discontinuity in the interval \((0,\pi)\).

Log in to reply

@parv mor

Log in to reply

@Rishabh Cool. I have been trying this one for sometime,

Log in to reply