Question

Reimann Sum... I got it wrong twice... Please help and provide step by step clear explaination. I'd appreciate it.

Answer 1

Formula for Riemann Sum is:

`(b-a)/(n) \sum_(i=1)^n f(a + i (b-a)/(n))`
interval is [1,3] so a = 1, b = 3
f(x) = 3x , sub into Riemann sum


`(2)/(n) \sum_(i=1)^n 3(1 + (2i)/(n))`

Continue by simplifying using properties of summations.

`= (2)/(n)\sum_(i=1)^n 3 + (2)/(n)\sum_(i=1)^n (6i)/(n) \\ \\ = (6)/(n)\sum_(i=1)^n 1 + (12)/(n^2)\sum_(i=1)^n i \\ \\ =(6)/(n) (n) + (12)/(n^2)((n(n+1))/(2)) \\ \\ =6+(6)/(n)(n+1) \\ \\ =12 + (6)/(n)`

Now you have an expression for the summation in terms of 'n'.

Next, take the limit as n-> infinity.
The limit of
`(6)/(n)` goes to 0, therefore the limit of the summation is 12.

The area under the curve from [1,3] is equal to limit of summation which is 12.