# Prove that the function $f(x)=x$ is integrable on $[-1,1]$ and $\int_{-1}^{1}x\,dx=0.$

Prove that the function $f(x)=x$ is integrable on $[-1,1]$ and $\displaystyle\int_{-1}^{1}x\,dx=0.$

My goal is to find a particular partition $P_n$ such that for every $\epsilon >0$ then $U(f,P_n)-L(f,P_n) < \epsilon$. Now my issue is how to find such a partition. I want to divide $[-1,1]$ into sub-intervals of a particular length. I figure there is something about the symmetry so having an event number of sub-intervals would be nice but not sure about the length.

• Continuous functions are regulated, which are integrable. To find the value of the integral, either use a suitable sequence of step functions or note that the integral is of an odd function over a symmetric interval which is easy to prove is 0 using a substitution. – PhysicsMathsLove Apr 24 '18 at 17:07
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The $P$ will need to depend on $\epsilon$; you will not be able to find one $P$ for all $\epsilon$. To put it another way, your quantifiers are in the wrong order.

That said, the key property enabling you to do this problem directly from the definition is that $f(x)=x$ is an increasing function. This means

$$U(f,P)=\sum_{i=1}^n f(x_i)(x_i-x_{i-1}) \\ L(f,P)=\sum_{i=1}^n f(x_{i-1})(x_i-x_{i-1})$$

where $P$ is $-1=x_0<x_1<\dots<x_n=1$. To make these close to each other, a uniform partition will suffice.

Hint. Take as $P_n$ the uniform partition $x_k=\frac{k}{n}$ with $k=-n,\dots,n$ where $n$ is a positive integer. Then, since $f(x)=x$ is strictly increasing, it follows that $$U(f,P_n)=\frac{1}{n}\sum_{k=-n+1}^n\frac{k}{n}\quad\mbox{and}\quad L(f,P_n)=\frac{1}{n}\sum_{k=-n}^{n-1}\frac{k}{n}.$$ Are you able to find $n$ such that $U(f,P_n)-L(f,P_n)<\epsilon$?

The function is continuous on this interval so it’s integrable by definition.

Using Riemann integral theory you can pick any Pn.

EDIT: I can relate to what’s been said in the comment but it depends on what equivalent definition you use.

• One does not usually define continuous functions to be integrable, it is a consequence of the integration theory. – Ian Apr 24 '18 at 16:49
• It is also clear from the context that the OP is either using the Darboux definition, or has proven that whatever other definition they are starting from is equivalent to the Darboux definition. Thus you can assume that they start from there. – Ian Apr 24 '18 at 16:55