Let $\displaystyle f(x)=2$ if $\displaystyle 0\leq x<1$ and $\displaystyle f(x)=1$ if $\displaystyle 1\leq x\leq 2$ . Show that $\displaystyle f\in R[0,2]$ and evaluate its integral.
I'm not sure of the latex code for partitions...
Let $\displaystyle f(x)=2$ if $\displaystyle 0\leq x<1$ and $\displaystyle f(x)=1$ if $\displaystyle 1\leq x\leq 2$ . Show that $\displaystyle f\in R[0,2]$ and evaluate its integral.
I'm not sure of the latex code for partitions...
Which part do you need help on? The value of the integral is pretty clear (it's just two boxes, so calculate the area of the boxes).
As for the integrability, try treating them as two functions $\displaystyle g_1(x)=2\cdot\mathbf{1}_{[0,1)}(x)$ and $\displaystyle g_2(x)=\mathbf{1}_{[1,2]}(x)$ where $\displaystyle \mathbf{1}_A(x)=1$ only when $\displaystyle x \in A$ and is zero otherwise.
When you show that both of those are integrable, then you know that the sum of them (which gives you f) is integrable.