Help: Continuity of function 2

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 $\text{\_\_\_\_\_\_\_\_\_\_}$ .

#Calculus

Easy Math Editor

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Comments

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 $(0,\pi)$ .

Abhishek Sinha - 5 years, 3 months ago

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

Akhilesh Prasad - 5 years, 4 months ago

@parv mor

Akhilesh Prasad - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$