## anonymous one year ago 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.

1. anonymous

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

2. anonymous

Same rules correct?

3. jim_thompson5910

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

4. anonymous

a=2 b=5

5. jim_thompson5910

good

6. jim_thompson5910

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

7. anonymous

It should be just x Final answer should be D

8. jim_thompson5910

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

9. anonymous
10. jim_thompson5910

yep it's D

11. anonymous

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

12. jim_thompson5910

sure

13. anonymous
14. jim_thompson5910

just as long as it doesn't get too cluttered

15. anonymous

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

16. jim_thompson5910

They're both correct. Nice job

17. anonymous

Appreciate it man

18. jim_thompson5910

no problem