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Homework Help: Definite Integrals

  1. Mar 6, 2010 #1
    Hi all, just wondering if someone could explain to me about define integrals.
    Say i have F(x)=0.5x2-2 and F(x)=x3+x2-6x and i want to find the area of the regions which is satisfied from -2 to 2.

    so [tex]\int[/tex]0.5x2-2 from -2 to 2. [tex]\int[/tex]x3+x2-6x from -2 to 2.

    Now with the cubic, from -2 to 0 is supposedly positive and from 0 to 2 it's negative, is that true i thought the it doesn't matter which region the area is in it'll always be counted as positive.

    so, [tex]\int[/tex]x3+x2-6x from -2 to 0. = [tex]\int[/tex](-2)3+(-2)2-6(-2) - [tex]\int[/tex](0)3+(0)2-6(0) = 8-0 = 8units2


    [tex]\int[/tex]x3+x2-6x from 0 to 2. = [tex]\int[/tex](0)3+(0)2-6(0) - [tex]\int[/tex](2)3+(2)2-6(2) = 0- 0 = 0 units2

    Therefore, Area from (-2 to 0)+ Area from (0 to 2) = 8+0 = 8units2

    But using my calculator doing the same steps (except in the graphing section) doing -2 to 0 and adding it to 0 to 2's value i get 16units2( 10.667+5.333)

    I don't know whats going on..
     
  2. jcsd
  3. Mar 6, 2010 #2
    I only did the one area for the 2nd function, ignore the 1st function.
     
  4. Mar 6, 2010 #3
    A definite integral is positive if the area is above the y-axis and negative if the area is below the y-axis. If you integrate over an interval that has both, the result will then not represent the complete physical area.

    Looking at a graph of your cubic function: http://www.wolframalpha.com/input/?i=x^3%2Bx^2-6x

    The area under the curve from -2 to 0 is larger than the area "above" the curve from 0 to 2. Since the positive area above (-2,0) is larger than the negative area below (0,2), the total area will be positive.

    When I did these calculations, I got an area of 32/3 on (-2,0) and -16/3 on (0,2). So the definite integral on (-2,2) would be 32/3 - 16/3 = 16/3.

    The total physical area would be 32/3 + 16/3 = 48/3 = 16. I had to use the absolute value of the areas to calculate the total physical area. I hope this clears things up!
     
  5. Mar 6, 2010 #4

    HallsofIvy

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    You haven't integrated anything, you have just evaluated the function itself at -2 and 0 and subtracted. The integral is
    [tex]\int_{-2}^0 x^3+ x^2- 6x dx= \left[\frac{1}{4}x^4+ \frac{1}{3}x^3- 3x^2\right]_{-2}^0[/tex]
    [tex]= 0- \left(\frac{1}{4}(16)+ \frac{1}{3}(-8)- 3(4)\right)= \frac{32}{3}[/tex]


    That's because your calculator integrated and you didn't!
     
  6. Mar 6, 2010 #5
    With both did you do F(b) - F(a) ? also, do you know anything about these type of intergrals.

    [tex]\int[/tex]f(x)dx+|[tex]\int[/tex]f(x)dx|
    |[tex]\int[/tex]f(x)dx|+[tex]\int[/tex]f(x)dx
    |[tex]\int[/tex]f(x)dx+[tex]\int[/tex]f(x)dx|

    I'm unsure what these | | parts do when theyre associated with the integral. thanks alot.
     
  7. Mar 6, 2010 #6

    HallsofIvy

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    [tex]\left|\int f(x)dx\right|[/tex]
    means "first take the integral of the function then take the absolute value of that number".

    [tex]\int |f(x)|dx[/tex]
    means "first take the absolute value of the function, then integrate".
     
  8. Mar 6, 2010 #7
    Wow, i'm so retarded I forgot to actually intergrate the function. how embarrassing..
     
  9. Mar 6, 2010 #8
    Thanks alot HallsofIvy, I don't really understand. could you please do an example of both?
     
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