Determine a region whose area is equal to the given limit.

lim x-> infinity Sigma (n on top, i=1 on bottom) (2/n)*[5+(2i/n)]^10

I started with delta x=b-a/n =5-0/n =5/n

The width of delta is 2/n so b-a=2 I'm not sure how you got 5-0/n
The integrand looks like (5+x)^10, the limits look like they might be 0 to 2 I'll let you verify if this is right.

I might be mistake on the limits, it could be x= -5 to -3. I haven't done Riemann sums in some time so I'm a little rusty on this.

Now I think I was completely off the mark. The function is f(x) x^10 and the limits are 5 to 7. Shows how out of practice I am!

I reviewed Riemann sums briefly to make sure I know how they're set up. If we let delta-x = dx =(b-a)/n then we have
lim n->infty Sum_i=1-to-n[f(a+i*dx)*dx]
Here a=5, and (b-a)/n = 2/n, b-a=2 so b=7.
As I mentioned, f(x)=x^10
This is the typical form for the sum using the left endpoint for the starting point and the right endpoint of each rectangle for the height, thus these would be circumscribed rectangles since the function is increasing on this interval. The way to determine what function the sum corresponds to is to determine what the dx part is: for this problem it's 2/n. Then determine what the value for a is: here it's 5. You then find b, then f(x).
I'm not sure if this problem is asking you to determine the area of the region or simply the region itself. If you had to determine the area it would be the defintite integral of x^10dx from x=5 to x=7. I'm sure you know how to do this part.
Hopefully this clarifies my previous posts. As I said, it's been awhile since I set these up and I couldn't seem to recall the basics. Looking at a couple sites brought it back fairly quickly though.

  1. 👍
  2. 👎
  3. 👁

Respond to this Question

First Name

Your Response

Similar Questions

  1. Help me check my calculus answers

    1. Which of the following functions grows the fastest as x goes to infinity? - 2^x - 3^x - e^x inf f(x)/g(x) = 5 show? - g(x) grows faster than f(x) as x goes to infinity. - f(x) and g(x) grow at the same rate as x goes to

  2. Calculus

    What does lim x-->∞ f(x)/g(x)=0 show? a) g(x) grows faster than f(x) as x goes to infinity b) f(g) and g(x) grow at the same rate as x goes to infinity c) f(x) grows faster than g(x) as x goes to infinity d) LHopital's Rule must

  3. Calculus check please

    1. Which of the following functions grows the fastest as x goes to infinity? - 2^x - 3^x - e^x (my answer) - x^20 2. Compare the rates of growth of f(x) = x + sinx and g(x) = x as x approaches infinity. - f(x) grows faster than

  4. Calculus

    Evaluate the limit. as x approaches infinity, lim sqrt(x^2+6x+3)-x

  1. Calculus

    The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles. A = lim n → ∞ [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx] Use this definition to

  2. Calculus, please check my answers!

    1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 The


    What is the following limit? lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n) II.) Definite integral from 0 to pi of

  4. calc bc (condensed

    is the limit as x approaches 0 of sin3x over 3x equal to zero? sorry-- basically this is my problem: lim [sin 3x / 4x) x-> 0 ~~~~I multiplied& eventually got to .75* lim (sin 3x / 3x) x-> 0 ~so i figured since (lim (sinx/x) x-> 0

  1. Calculus Limits

    Question: If lim(f(x)/x)=-5 as x approaches 0, then lim(x^2(f(-1/x^2))) as x approaches infinity is equal to (a) 5 (b) -5 (c) -infinity (d) 1/5 (e) none of these The answer key says (a) 5. So this is what I know: Since

  2. Single Variable Calculus

    find an expression for the area under the graph of f(x)= (x^2)+x from x=2 to x=5 as a limit of a riemann sum (do not need to evaluate). the answer i got was: lim as x-> infinity of sigma from i=2 to n of (2+3i/n)^2+(3i/n)(3/n) is

  3. Calculus

    Evaluate the limit using L'Hospital's rule if necessary. lim as x goes to +infinity x^(6/x)

  4. Pre-cal

    Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity lim 4n-3 / 3n^2+2 n-> infinity I did lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity (2+6x-3x²)/(4x²+4x+1)

You can view more similar questions or ask a new question.