Integration using the limit definition: Why does f(2+(4i/n))(4/n) become 8(4/n)?

[gsrokmix]

Hi,

I am having a terrible time understanding why f(2 + (4i/n))(4/n) becomes 8(4/n). Can someone please explain what happens between these 2 steps? Wouldn't (4/n) go outside the sum, and then what happens to the 2 being added? Thanks so very much - George


\(\displaystyle 3.\quad y\, =\, 8\, \mbox{ on }\, [2,\, 6].\)

. . .\(\displaystyle \big(\mbox{Note: }\, \Delta x\, =\, \dfrac{6\, -\, 2}{n}\, =\, \dfrac{4}{n},\, \Vert \Delta \Vert \, \rightarrow\, 0\, \mbox{ as }\, n\, \rightarrow\, \infty \big)\)

. . .$\displaystyle \displaystyle \sum_{i = 1}^n\, f(c_i)\, \Delta x_i\, =\, \sum_{i = 1}^n\, f\left(2\, +\, \dfrac{4i}{n}\right)\left(\dfrac{4}{n}\right)$

. . . . . .$\displaystyle \displaystyle =\, \sum_{i = 1}^n\, 8\, \left(\dfrac{4}{n}\right)\, =\, \sum_{i = 1}^n\, \dfrac{32}{n}\, =\, \dfrac{1}{n}\, \sum_{i = 1}^n\, 32$

. . . . . . . . .$\displaystyle =\, \dfrac{1}{n}\, (32n)\, =\, 32$

. . . . . .$\displaystyle \displaystyle \int_2^6\, 8\, dx\, =\, \lim_{n \rightarrow \infty}\, 32\, =\, 32$

 

[ksdhart2]

The $\displaystyle f \left( 2+\dfrac{4i}{n} \right)$ term becomes 8, because that's how the function you're integrating is defined. You're integrating $\displaystyle y=f(x)=8$, so for any x value, in this case $\displaystyle x = 2 + \dfrac{4i}{n}$, it always equals 8.

 

[gsrokmix]

Ahhh, yes I forgot! Thanks very much!
- George