# Riemann integrability

• Dec 6th 2009, 06:07 PM
thaopanda
Riemann integrability
Decide which of the functions $f_n$ : [0,1] $\rightarrow$ R with n = 0,1,2, defined by setting:

$f_n(x)$ :=
$(x^n)sin(\frac{1}{2}$ if $0 < x \leq 1$
0 if x = 0

are Riemann integrable on [0,1].

• Dec 6th 2009, 09:54 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
Decide which of the functions $f_n$ : [0,1] $\rightarrow$ R with n = 0,1,2, defined by setting:

$f_n(x)$ :=
$(x^n)sin(\frac{1}{2}$ if $0 < x \leq 1$
0 if x = 0

are Riemann integrable on [0,1].

Is that $f_n(x)=x^n\sin\left(\frac{1}{2}\right)
$
?
• Dec 7th 2009, 05:46 AM
thaopanda
yeah, sorry, I forgot the other parenthesis
• Dec 7th 2009, 07:40 AM
redsoxfan325
Quote:

Originally Posted by thaopanda
yeah, sorry, I forgot the other parenthesis

Well $\sin(1/2)$ is just a number. Call it $k$. So you need to know for which n is $f_n(x)=kx^n$ integrable on $[0,1]$.

For $n\geq1$, it's continuous and therefore integrable. For $n=0$, you have one removable discontinuity (at $x=0$), so it's also integrable.
• Dec 7th 2009, 07:49 AM
thaopanda
if it was $sin(\frac{1}{x})$, how would that change the problem?
• Dec 7th 2009, 12:12 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
if it was $sin(\frac{1}{x})$, how would that change the problem?

$x^n\sin(1/x)$ is continuous for all $n\geq1$. If $n=0$, the function is still continuous on $(0,1]$ and is therefore integrable.