1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the area bounded

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the area bounded by y = 3x^2 + 1, x = 0, x = 2, y = 0


    2. Relevant equations



    3. The attempt at a solution
    Not sure how to do this? Is this like finding the upper and lower sums?
     
  2. jcsd
  3. Nov 25, 2007 #2
    >Find the area bounded by y = 3x^2 + 1, x = 0, x = 2, y = 0
    >Not sure how to do this? Is this like finding the upper and lower sums?

    If you draw a picture, it is quickly seen that the area is the integral of f(x) = 3x^2 from 0 to 2: \int_0^2 3x^2 dx.

    Using upper and lower sums:
    You would have to calculate the limit of the upper sum as the interval is divided to smaller and smaller subintervals. Then you should do the same for the lower sum and conclude that because the two limits are equal, the integral (=area) must also be equal to this number.
     
  4. Nov 25, 2007 #3
    So I can find the area of the upper and lower sum by using like a change in X of 1/4? Because in order for them to equal I would have to be using a change in X of like 1/100 or I don't know but something very small right?
     
  5. Nov 25, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, no specific value for that will give you the exact area. That's why Mikko said "limit".

    One of the reasons you learn about "upper and lower sums" (more generally Riemann sums) is to help you set up integrals. If you were to divide that area into rectangles you could approximate the area as the sum of those rectangles. If you divide the interval from x= 0 to x=2 into "n" equal parts so that each part has length 2/n, then
    [itex]A= \sum_{i=1}^{n} (3(2i/n)^2+ 1)(2/n)= \frac{24}{n}\sum_{i=1}^{n}i^2+ (2/n)\sum_{i=1}^{n}1= \frac{24}{n^3}\sum_{i=1}^{n}i^2+ 2[/tex]. If you are very clever and can recall a formula for [itex]\sum i^2[/itex] you can get that as a function of n and then take the limit as n goes to infinity.

    Of course, you can see that that is very tedious to do in general! Do you know about "anti-derivatives"?
     
  6. Nov 25, 2007 #5

    PhY

    User Avatar

    I think the Area is 10 Square Units with or without the y=0.

    I don't know how to integrate 0.

    Working:
    1 bounded area +ve
    A= 2[int]0 ydx
    A=2[int]0 3x^2 +1 dx
    A= [x^3 + x]2/0
    A=[2^3+2^1]-[0^3+0^1]
    A=10 Squareunits

    Problem with this is it covers the -negative area aswell when the y=0 should stop it.
    Can't find anything in my notes.Can't integrate zero.
     
  7. Nov 25, 2007 #6

    rock.freak667

    User Avatar
    Homework Helper

    Well...if you were to sketch the curve, you'd see that y=0 is the x-axis and x=0 is the y-axis
    and that the area is simply given by

    [tex]\int_0 ^{2} (3x^2+1)dx[/tex]
     
  8. Nov 25, 2007 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ?? What happened to " BuBbLeS01". I don't know what you mean by "without the y= 0"! Without some lower boundary you don't have a region to find the area of. The upper boundary is given y= 3x2+ 1 and the lower boundary by y= 0 so a "thin rectangle" would have length 3x2+ 1- 0= x2+ 1, and width dx. That is what you integrate (from 0 to 2 not from 2 to 0).

    I can "integrate 0": the integral of any constant is that constant time x: [itex]\int 0 dx= 0+ C[/itex].
     
  9. Nov 25, 2007 #8
    Ok so can I use the way RockFreak set it up? I do know about anti-derivatives. So do I find the anti-derivative of 3x^2 + 1? Then plug in 2 and 0 and subtract them?

    Anti-derivative of 3x^2 + 1 = x^3 + x
    Plug in 2 and 0 =(2^3 + 2) - (0^3 + 0) = 10
     
  10. Nov 25, 2007 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's right. Although it's a lot more fun to set up the Riemann sums, don't you think!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?