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Help me check my HW (area under a curve)

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    I need to use the right endpoint formula to find the area under the curve f(x) = 4-x2 over the interval [1,2]
    I integrated this problem and got 5/3. But doing it the long way got me 11/3. Can someone where I made my mistake on applying the right endpoint formula, thanks!



    2. Relevant equations
    right endpoint formula = [tex] f(\frac{b-a}{n}*k) [/tex] and [tex]
    \sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}[/tex]



    3. The attempt at a solution
    Before I use the right endpoint formula I tried integration and got:
    [tex]\int_{1}^{2} 4-x^2 dx = \left[ 4x-\frac{x^3}{3}\right]_{1}^{2} = \frac{5}{3}[/tex]

    here is what I got using the right endpoint formula:
    [tex] \lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n}
    = \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n} [/tex]
    [tex] = \frac{4}{n} - \frac{K^2}{n^3} [/tex]
    [tex] \lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )[/tex]
    [tex] = 4 - \frac{1}{3} = \frac{11}{3} [/tex]
     
  2. jcsd
  3. Mar 28, 2010 #2
    You made a mistake using the right endpoint, mainly the set up of the problem the way I see it.

    It should be something like:

    lim as n --> infinity

    summation from k = 1 to n of

    [4 - (1 + k(1/n))^2] x (1/n)
     
  4. Mar 28, 2010 #3
    Hint: try evaluating [tex]\int_0^1 f(x) dx[/tex]
     
  5. Mar 28, 2010 #4
    I looked at my setup again, but I still don't see the mistake. Where did you get the 1 from? [4 - (1 + k(1/n))^2] can you explain a little more please?
    Thanks
     
  6. Mar 30, 2010 #5
    what you substitute for x should be in the form a + k(delta x) since it's a right endpoint approximation (at least the way I learned it). where a is the lower bound of integration or first number in the interval, in this case 1.

    Edit: I get 5/3 my way.
     
    Last edited: Mar 30, 2010
  7. Mar 30, 2010 #6
    Yea, the formula I used is off that's why I was not able to figure out which part I made a mistake on.

    While on the subject of formula, do I switch formula when I have an increasing function?
     
  8. Mar 30, 2010 #7
    What do you mean?

    The formula depends on where you are approximating, i.e. right endpoint, left endpoint or midpoint.
     
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