Partitioning the interval \([6,11]\) into \(n\) equally-spaced subintervals gives \(n\) rectangles of width \(\Delta x=\dfrac{11-6}n=\dfrac5n\) and of heights determined by the right endpoints of each subinterval.
If \(a=x_1=6\), then \(x_2=6+\dfrac5n\), \(x_3=6+2\dfrac5n\), and so on, up to \(b=x_{n+1}=6+n\dfrac5n=11\). Because we're using the right endpoints, the approximation will consider \(x_2,\ldots,x_{n+1}\)
The definite integral is then approximated by
\(\displaystyle\int_6^{11}(1-5x)\,\mathrm dx\approx\sum_{i=2}^{n+1}f(x_i)\Delta x=\sum_{i=2}^{n+1}\left(1-\left(6+(i-1)\dfrac5n\right)\dfrac5n\)
You have
\(\displaystyle\sum_{i=2}^{n+1}\left(1-5\left(6+(i-1)\dfrac5n\right)\dfrac5n=\sum_{i=2}^{n+1}\left(-29-\dfrac{25}n(i-1)\right)\dfrac5n\)
\(=\displaystyle-{145}n\sum_{i=2}^{n+1}1-\dfrac{125}{n^2}\sum_{i=2}^{n+1}(i-1)\)
\(=-145-\dfrac{125(n^2+n)}{2n^2}\)
\(=-\dfrac{145}2-\dfrac{125}{2n}\)
To check that this is correct, let's make sure the sum converges to the exact value of the definite integral. As \(n\to\infty\), you have the sum converging to \(-\dfrac{145}2\).
Meanwhile,
\(\int_6^{11}(1-5x)\,\mathrm dx=\left[x-\dfrac52x^2\right]_{x=6}^{x=11}=-\dfrac{415}2\)
so we're done.
