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

Question

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.

Answer 1

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.

Answer 2

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 $U(f,P_n)-L(f,P_n)<\epsilon$?

Answer 3

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.