# Homework Help: Need some help with Riemann Sums.

1. Dec 1, 2008

### Kuma

Need some urgent help with Riemann Sums.

1. The problem statement, all variables and given/known data

PART A:

In all of this question, let I = $$\int ^{2}_{-2} f(x)dx$$ $$where f(x) = -2x + 1$$

Evaluate I.

PART B:

Use the defintion of the definite integral to evaluate I.
i.e Riemann Sum.

2. Relevant equations

3. The attempt at a solution

PART A:

I = [-(2)^2 + (2)] - [-(-2)^2 + (-2)]
I = -2 + 6 = 4

Part B is what i am having trouble with.

Xk = a + k($$\Delta x$$)
a = -2

and

$$\Delta x$$ = b-a/n = 2- (-2)/n = 4/n

so then

Xk = -2 + 4k/n

Now the Reimann sum is:

$$\sum f(Xk)(\Delta x)$$ = $$\sum [-2(-2 + 4k/n) + 1](4/n)$$
= 4/n $$\sum 4 - 8k/n + 1$$
= 4/n $$\sum -8k/n + 5$$
= 4/n x -8/n $$\sum k$$ + $$\sum 5$$
= -32/n2 x [(n)(n+1)/2] + 5n

Now when i take the limit approaching infinity for that, it doesnt give me 4 as an answer unless im doing something wrong.
This is where i am stuck at.

2. Dec 1, 2008

### Dick

Re: Need some urgent help with Riemann Sums.

Your last two lines start to look pretty goofy. The limit of (4/n)*(5n-8*n(n+1)/2) is 4. Then parentheses start disappearing and numbers are magically factored out. Check it again.