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.

Thank you in advance!

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 $\varepsilon$.

  • 1
    $\begingroup$ math.stackexchange.com/questions/2715623/… $\endgroup$ – user284331 Apr 6 '18 at 0:28
  • $\begingroup$ I tried different attempt with your reference, but I still get the answer , which is larger than $\varepsilon$. Could you check it? $\endgroup$ – Si Hyun Kim Apr 6 '18 at 1:21
  • $\begingroup$ What do you mean by larger than $\epsilon$? Is that $2\epsilon$ or what? $\endgroup$ – user284331 Apr 6 '18 at 1:26
  • $\begingroup$ 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. $\endgroup$ – 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 $\displaystyle\int_{0}^{1}f(x)dx=\dfrac{1}{2}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.