Need some help with Riemann Sums.

[Kuma]

Need some urgent help with Riemann Sums.

Homework Statement

PART A:

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

Evaluate I.


PART B:

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

Homework Equations

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.

X_k = a + k([tex]\Delta x[/tex])
a = -2

and

[tex]\Delta x[/tex] = b-a/n = 2- (-2)/n = 4/n

so then

X_k = -2 + 4k/n

Now the Reimann sum is:

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

Now when i take the limit approaching infinity for that, it doesn't give me 4 as an answer unless I am doing something wrong.
This is where i am stuck at.

 

[Dick]

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.