# Show that $f\in \mathscr{R}[0,1]$ and compute $\int_0^1f$ using the definition of the integral

Question: Let $f:[0,1]\rightarrow R$ be defined by $f(x)=x$. Show that $f\in \mathscr{R}[0,1]$ and compute $\int_0^1f$ using the definition of the integral (but feel free touse the propositions of this section).

Attempt: The definition of Riemann integral is that for every $\varepsilon>0$, there exists a partition $P$ such that $U(P,f)-L(P,f)<\varepsilon$.

I claim that $P=\{0,\varepsilon,1-\varepsilon,1\}$, and compute $m_i=\inf\{f(x):x_{i-1}\le x \le x_i\}$ for $i=1,2,3$. Similarly, I compute $M_i=\sup\{f(x):x_{i-1}\le x \le x_i\}$. Then, $L(P,f)=\sum_{i=1}^{3}\Delta x_im_i$, and $U(P,f)=\sum_{i=1}^{3}\Delta x_iM_i$. Then, I subtract these two, but I get the answer which is larger than $\varepsilon$. So, my question is how can we first determine a partition $P$, which satisfies the definition of the integral to solve this type of question.

Edit: We know $m_i=x_{i-1}$ and $M_i=x_i$ since $f(x)=x$. Then, $L(P,f)=\sum_{i=1}^{n}\Delta x_im_i=\sum_{i=1}^{n} (x_i-x_{i-1})x_{i-1}=(x_n-x_0)\sum_{i=1}^{n}x_{i-1}$.

$U(P,f)=\sum_{i=1}^{n}\Delta x_iM_i=\sum_{i=1}^{n}(x_i-x_{i-1})x_i=(x_n-x_0)\sum_{i=1}^{n}x_i$.

Then, $U(P,f)-L(P,f)=(x_n-x_0)\sum_{i=1}^{n}x_i-(x_n-x_0)\sum_{i=1}^{n}x_{i-1}=(x_n-x_0)(x_n-x_0)=1$. But, this is larger than $\varepsilon$.

• math.stackexchange.com/questions/2715623/… – user284331 Apr 6 '18 at 0:28
• I tried different attempt with your reference, but I still get the answer , which is larger than $\varepsilon$. Could you check it? – Si Hyun Kim Apr 6 '18 at 1:21
• What do you mean by larger than $\epsilon$? Is that $2\epsilon$ or what? – user284331 Apr 6 '18 at 1:26
• To satisfy the definition of the integral, $U(P,f)-L(P,f)<\varepsilon$, but I got $U(P,f)-L(P,f)=1$. Since $\varepsilon>0$ is arbitrary, $\varepsilon<1$, which does not satisfy the definition. – Si Hyun Kim Apr 6 '18 at 1:30

That $\displaystyle\sum_{i=1}^{n}(x_{i}-x_{i-1})x_{i-1}=(x_{n}-x_{0})\displaystyle\sum_{i=1}^{n}x_{i-1}$ is not valid, that $\displaystyle\sum u_{i}v_{i}=\sum u_{i}\sum v_{i}$ is not a valid formula, it is easy to come with counterexamples.
Rather, it is proved in my post that given $\epsilon>0$, one can find a partition $P$ on $[0,1]$ such that $U(P,f)\leq\dfrac{1}{2}+\epsilon$, so $\inf_{P}U(P,f)\leq\dfrac{1}{2}$. It is also proved that $U(f,P)\geq\dfrac{1}{2}$ for any partition $P$, so $\inf_{P}U(P,f)=\dfrac{1}{2}$.
Once we can prove that $U(P,f)-L(P,f)<\epsilon$ for a partition $P$, then $f$ is Riemann integrable, and the integral is $\dfrac{1}{2}$ because of that $\inf_{P}U(P,f)=\dfrac{1}{2}$. Note that for the partition $P=\{0,1/n,2/n,...,n/n\}$ such that $1/n<\epsilon$, then \begin{align*} U(P,f)-L(P,f)&=\sum_{i=1}^{n}(x_{i}+x_{i-1})(x_{i}-x_{i-1})\\ &=\sum_{i=1}^{n}(x_{i}^{2}-x_{i-1}^{2})\\ &=x_{n}^{2}-x_{0}^{2}\\ &=\dfrac{1}{n^{2}}\\ &<\dfrac{1}{n}\\ &<\epsilon, \end{align*} we are done.
For the other approach, note that \begin{align*} L(P,f)&=\sum_{i=1}^{n}m_{i}\Delta x_{i}\\ &=\sum_{i=1}^{n}x_{i-1}(x_{i}-x_{i-1})\\ &<\sum_{i=1}^{n}\dfrac{1}{2}(x_{i}+x_{i-1})(x_{i}-x_{i-1})\\ &=\dfrac{1}{2}(x_{n}^{2}-x_{0}^{2})\\ &=\dfrac{1}{2}, \end{align*} so $\sup_{P}L(P,f)\leq\dfrac{1}{2}$. Given $\epsilon>0$, choose the partition $P=\{0,1/n,2/n,...,n/n\}$ such that $1/n<\epsilon$, then \begin{align*} L(P,f)&=\sum_{i=1}^{n}\dfrac{i-1}{n}\cdot\dfrac{1}{n}\\ &=\dfrac{1}{n^{2}}\sum_{i=1}^{n}(i-1)\\ &=\dfrac{1}{n^{2}}\cdot\left(\dfrac{n(n+1)}{2}-n\right)\\ &=\dfrac{1}{2}-\dfrac{1}{2n}\\ &>\dfrac{1}{2}-\dfrac{1}{n}\\ &>\dfrac{1}{2}-\epsilon, \end{align*} so $\sup_{P}L(P,f)\geq\dfrac{1}{2}$, and hence $\sup_{P}L(P,f)=\dfrac{1}{2}$, we conclude that $\inf_{P}U(P,f)=\sup_{P}L(P,f)=\dfrac{1}{2}$, so $\displaystyle\int_{0}^{1}f(x)dx=\dfrac{1}{2}$.