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.
Same rules correct?
do you see what the value of 'a' is?
a=2 b=5
good
now do you see what f(x) must be?
It should be just x Final answer should be D
I can't see your answer choices, but you'll find that a = 2 b = 5 f(x) = x so \[\Large \lim_{n \to \infty}\sum_{i=1}^{n}\left(2+\frac{3i}{n}\right)*\frac{3}{n} = \int_{2}^{5}x dx\]
yep it's D
Is it cool if I just post two other questions here so you can just check them? Instead of making other questions?
sure
https://gyazo.com/b78cff91955b504f923a0d1e029f56dd https://gyazo.com/af655f0b48727eebe89ad8bbfec61f24
just as long as it doesn't get too cluttered
If either one is wrong, I'll make a new question.
They're both correct. Nice job
Appreciate it man
no problem
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