Show: $\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx=\sum_{m=1}^{k}m \int_m^{m+1}f(x)dx$ What I understand is that $\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx$  means summing $\int_m^{m+1}f(x)dx$ from $1$ to $k$ twice, since we see that  $\sum_{n=1}^{k}=\sum_{m=1}^{k}$ since, $m=n$, but then how we get the identity given above? I am confused!
Please show elaborately that- 
$$\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx$$
$$=\sum_{m=1}^{k}m \int_m^{m+1}f(x)dx$$
Thanks.

It is just a matter of definition.
$$
\sum_{n=1}^{k} \sum_{m=n}^{k} \int_m^{m+1} f(x)dx
$$
Expand along the first sum:
$$
\left( \sum_{m=1}^{k} \int_m^{m+1} f(x)dx \right)
+ \left( \sum_{m=2}^{k} \int_m^{m+1} f(x)dx \right)
+ \cdots
+ \left( \sum_{m=k}^{k} \int_m^{m+1} f(x)dx \right)
$$
Write each term in parentheses as a new line, and expand the sum:
$$
\begin{array}{ccccccc}
\int_1^{2} f(x)dx & + & \int_2^{3} f(x)dx & + & \cdots & + & \int_k^{k+1} f(x)dx \\
& + & \int_2^{3} f(x)dx & + & \cdots & + & \int_k^{k+1} f(x)dx \\
&&&& \ddots && \vdots \\
&&&&& + & \int_k^{k+1} f(x)dx
\end{array}
$$
Now count the number of each integral:
$$
1 \cdot \left( \int_1^{2} f(x)dx \right)
+ 2 \cdot \left( \int_2^{3} f(x)dx \right)
+ \cdots
+ k \cdot \left( \int_k^{k+1} f(x)dx \right)
$$
Recombine using sigma notation:
$$
\sum_{m=1}^{k} m \int_m^{m+1} f(x)dx
$$
