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 .

Question

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 1 plus 3 times i over n and 3 over n . (4 points)

Question 1 options:

1)

the integral from 1 to 3 of the quantity x plus 1, dx


2)

the integral from 1 to 4 of x, dx


3)

the integral from 1 to 4 of the quantity 3 times x plus 1, dx


4)

the integral from 1 to 3 of x, dx

Answer 1

All those words! It appears you have

∑ (1 + 3i/n)(3/n)

This seems to be a right-hand sum on an interval of width 3, since it is

∑ (1+(3/n)i)(3/n) = ∑f(xi)∆x

That makes me think that it is

∫[1,4] x dx

If you have learned to evaluate definite integrals, you should verify that the sum and the integral are equal.

Answer 2

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 squared of 3 plus 5 times i over n and 5 over n.

the integral from 3 to 8 of x squared, dx

the integral from 0 to 3 of x plus 3 quantity squared, dx

the integral from 0 to 5 of the quantity 3 plus 5 times x, dx

the integral from 3 to 5 of x squared, dx