Let $f : [1,2] → \mathbb {R}$ be defined by $f(x) = x$. Prove that $f$ is Riemann integrable and compute $\int_1^2f(x)dx$ 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  $$\int_1^2f(x)dx=\lim_{n\to\infty}L(f,P_n)=\lim_{n\to\infty}U(f,P_n)=\frac32$$

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