# Riemann integrability

• Dec 6th 2009, 05:07 PM
thaopanda
Riemann integrability
Decide which of the functions $\displaystyle f_n$ : [0,1] $\displaystyle \rightarrow$ R with n = 0,1,2, defined by setting:

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

are Riemann integrable on [0,1].

more help please (Lipssealed)
• Dec 6th 2009, 08:54 PM
redsoxfan325
Quote:

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

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

are Riemann integrable on [0,1].

more help please (Lipssealed)

Is that $\displaystyle f_n(x)=x^n\sin\left(\frac{1}{2}\right)$ ?
• Dec 7th 2009, 04:46 AM
thaopanda
yeah, sorry, I forgot the other parenthesis
• Dec 7th 2009, 06:40 AM
redsoxfan325
Quote:

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

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

For $\displaystyle n\geq1$, it's continuous and therefore integrable. For $\displaystyle n=0$, you have one removable discontinuity (at $\displaystyle x=0$), so it's also integrable.
• Dec 7th 2009, 06:49 AM
thaopanda
if it was $\displaystyle sin(\frac{1}{x})$, how would that change the problem?
• Dec 7th 2009, 11:12 AM
redsoxfan325
Quote:

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

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