Integration involving d[x] instead of dx Find the value of: $$\int_0^3(x^2+1)d\lfloor x\rfloor$$
Shouldn't the answer be zero since $[x]$ is always an integer? But the options given are $12,17,15,19$. ([x] denotes floor(x))
From $0$ to $1$, $[x]=0$ and $d[x]=0$. Similarly, from $1$ to $2$, $[x]=2$ and $d[x]=0$ and so on...
 A: Let $f(x) = x^2+1$, $\alpha(x) = \lfloor x \rfloor$, and let $P$ be a partition of $[0,3]$ such that the points $1,2,3$ are in different partitions.
Let $I_k$ be the intervals in the partition and without loss of generality, suppose $k \in I_k$ for $k = 1,2,3$. Note that we have
$\alpha(\sup I_k) - \alpha(\inf I_k) = 1$ for $k=1,2,3$.
Also note that
on any other interval $J$ in the
partition, we have $\alpha(\sup J) - \alpha(\inf J) = 0$.
Hence we have
$L(P,f,\alpha) = \sum_{k=1}^3 \inf_{x \in I_k} f(x)(\alpha(\sup I_k) - \alpha(\inf I_k)) = \sum_{k=1}^3 \inf_{x \in I_k} f(x)$,
and similarly we have
$U(P,f,\alpha) = \sum_{k=1}^3 \sup_{x \in I_k} f(x)$.
Since $k \in I_k$ we see that
$L(P,f,\alpha) \le \sum_{k=1}^3 f(k)$ and similarly 
$U(P,f,\alpha) \ge \sum_{k=1}^3 f(k)$.
Since $f$ is continuous, for any $\epsilon>0$ we can choose a partition so that
$L(P,f,\alpha) \ge \sum_{k=1}^3 f(k)-\epsilon$,
$U(P,f,\alpha) \le \sum_{k=1}^3 f(k)+\epsilon$,
from which it follows that
$\int_0^3 f(x) d \alpha(x) = \sum_{k=1}^3 f(k)$.
Evaluating for the specific $f$ gives $17$.
