Let $f : [1,2] → \mathbb {R}$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and compute $$\int_1^2f(x)dx$$ as the limit of upper (and lower) sums.

I tried solving this but I don't know if my answer is right.

Let $P_n$ be the uniform partition of $[1,2]$ given by $$1 \lt 1+\frac1{n} \lt 1+\frac2{n} \lt \cdots \lt 1+\frac{n-1}{n}\lt2$$

The function $f(x)=x$ is increasing, hence $$m_i=inf_{x \in [x_{i-1},x_i]}f(x)=f(x_{i-1})=1+\frac{i-1}{n}$$ and $$M_i=sup_{x \in [x_{i-1},x_i]}f(x)=f(x_i)=1+\frac{i}{n}$$

$$L(f,P)=\sum_1^n(1+\frac{i-1}{n})\frac{i}{n}=\frac{3n-1}{2n}$$ and $$U(f,P)=\sum_1^n(1+\frac{i}{n})\frac{i}{n}=\frac{3n+1}{2n}$$

Therefore $$\lim_{n\to\infty}L(f,P_n)=\frac32 \quad \lim_{n\to\infty}U(f,P_n)=\frac32$$

By the Criterion of Integrability $$\int_1^2f(x)dx=\lim_{n\to\infty}L(f,P_n)=\lim_{n\to\infty}U(f,P_n)=\frac32$$

  • $\begingroup$ The function is continuous then is Riemann integrable. $\endgroup$ – Nosrati Aug 16 '18 at 9:55
  • $\begingroup$ yes but I had to prove as the limit of upper and lower sums $\endgroup$ – J.Dane Aug 16 '18 at 9:58

Yes, your proof is fine, everything is o.k.

  • $\begingroup$ Ok thanks, I wasn't sure when I took the values of $m_i$ and $M_i$ $\endgroup$ – J.Dane Aug 16 '18 at 9:49
  • $\begingroup$ Why the downvotes ????????????? $\endgroup$ – Fred Aug 16 '18 at 9:53
  • $\begingroup$ The OP wrote: " I don't know if my answer is right." I said: "your answer (proof) is fine". Why the dowmvotes ? $\endgroup$ – Fred Aug 16 '18 at 9:54
  • $\begingroup$ Is it really sufficient to consider uniform partitions only? $\endgroup$ – Martin R Aug 16 '18 at 10:09
  • $\begingroup$ It is sufficient: we have $|U(f,P_n)-L(f,P_n)|= \frac{1}{n}$. If $ \epsilon >0$ the choose $n$ such that $ \frac{1}{n}< \epsilon$. Riemann's criterion shows then that $f$ is R- integrable. It follows then that $U(f,P_n) \to \int_1^2f(x)dx$. $\endgroup$ – Fred Aug 16 '18 at 10:14

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