Question

I posted this question before, and unfortunately, I have to do it again because I just do not understand. It is so frustrating...

(see attached)

Answer 1

Hey blockcolder. You just reminded me to give you the "Best Answer" for the last question.

Answer 2

Any ideas?

Answer 3

I'm assuming this is a right hand sum.
If we compare your summation to the form of the Riemann sum, we can see that
\[\large a+i\Delta x=-5+i\frac{28}{n}\\
\large a=-5; \Delta x=\frac{28}{n}\\
\large b-a=28 \Rightarrow b=23\]

Answer 4

Yes. But then what do I do with the outer portions?

Answer 5

ie. 7i*14/n?

Answer 6

Manipulate it like this:
\[\large\frac{7i\cdot14}{n^2}=\frac{7i\cdot28}{2\cdot n^2}=\frac{7}{2n} \cdot\frac{28i}{n}=\frac{7}{2n}\cdot (x_i+5)\]
where x_i=-5+28i/n
Are you sure that's really supposed to be an n^2?

Answer 7

Yes that is an n^2 for sure.

Answer 8

Oh right. Just realized I rewrote it the wrong way:
\[\large \frac{7i \cdot 14}{n^2}=\frac{7i \cdot 28}{2n^2}=\frac{7}{2} \cdot \frac{i}{n} \cdot \frac{28}{n}=\frac{7}{2}\cdot \frac{x_i+5}{28}\cdot \Delta x\]
Thus, that horrible summation can be expressed as
\[\LARGE \int_{-5}^{23}\frac{7}{56}(x+5)f(x)\ dx\]