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- anonymous

The Riemann sum, the limit as the maximum of delta x sub i goes to infinity of the summation from i equals 1 to n of f of the quantity x star sub i times delta x sub i , is equivalent to the limit as n goes to infinity of the summation from i equals 1 to n of f of the quantity a plus i times delta x, times delta x with delta x equals the quotient of the quantity b minus a and n.
Write the integral that produces the same value as the limit as n goes to infinity of the summation from i equals 1 to n of the product of the quantity 2 plus 3 times i over n and 3 over n.

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- anonymous

- chestercat

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- anonymous

https://gyazo.com/4a2702de2f727a0265591e6cfd66d3a9
@jim_thompson5910

- anonymous

Same rules correct?

- jim_thompson5910

do you see what the value of 'a' is?

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- anonymous

a=2 b=5

- jim_thompson5910

good

- jim_thompson5910

now do you see what f(x) must be?

- anonymous

It should be just x
Final answer should be D

- jim_thompson5910

I can't see your answer choices, but you'll find that
a = 2
b = 5
f(x) = x
so
\[\Large \lim_{n \to \infty}\sum_{i=1}^{n}\left(2+\frac{3i}{n}\right)*\frac{3}{n} = \int_{2}^{5}x dx\]

- anonymous

https://gyazo.com/f9c6a5061ff442c9b97d4c357dd71289
:)

- jim_thompson5910

yep it's D

- anonymous

Is it cool if I just post two other questions here so you can just check them? Instead of making other questions?

- jim_thompson5910

sure

- anonymous

https://gyazo.com/b78cff91955b504f923a0d1e029f56dd
https://gyazo.com/af655f0b48727eebe89ad8bbfec61f24

- jim_thompson5910

just as long as it doesn't get too cluttered

- anonymous

If either one is wrong, I'll make a new question.

- jim_thompson5910

They're both correct. Nice job

- anonymous

Appreciate it man

- jim_thompson5910

no problem

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