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Homework Help: Cal Ii

  1. Jan 20, 2005 #1
    We're doing a review this week and i'm already having problems. I dont remeber doing this problem and cannot find similar problems in the book. Here's the question:

    You are given the three points in the plane [tex]A= (6-3)[/tex], [tex]B = (11,8)[/tex], and [tex]C=(15,0)[/tex]. The graph of the function [tex]f(x)[/tex] consists of the two line segments [tex]AB[/tex] and [tex]BC[/tex]. Find the integral [tex]\int_6^{15} f(x)dx[/tex] by interpreting the integral in terms of sums and/or differences of areas of elementary figures.

    1.) [tex]\int_6^{15} f(x)dx[/tex] = _______


    Here's what i done:

    the first thing i did was plugged in A,B, & C points into a graph. after that, i connected the dots and found out that there were two triangles. So i used the area of the triangle formula to find the area of each triangles and added it together. I got an answer of 43.5, but it's incorrect. I'm not good at math, so can someone point me to the right direction?
     
  2. jcsd
  3. Jan 20, 2005 #2

    dextercioby

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    Sorry,but i haven't made the calculations...But are u sure you added the areas with the right signs??The one above the Ox axis is positive,but the one below has a negative area...

    Daniel.
     
  4. Jan 20, 2005 #3
    What the integral does is add up all the area under the line ABC and "ABOVE THE X AXIS" .. you should have 2 triangles and 1 rectangle...... you really need a graph to do this problem at your level....
     
  5. Jan 20, 2005 #4
    wait... is that a A=(6,3) or A=(6,-3)... this makes a huge different...if it is a -3, you have to sum up all area above the x axis and minis the below the x axis.....
     
  6. Jan 20, 2005 #5

    dextercioby

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    I bet it is [tex] A(6,-3) [/tex].Else,he wouldn't have screwed up the calculations.

    Daniel.
     
  7. Jan 20, 2005 #6
    You may also wright analitically equation for straight line, which goes through 2 points.
    If P1 (x1, y1) and P2 (x2, y2), then

    (x-x1)/(x2-x1)=(y-y1)/(y2-y1) is the equation. And now You just have to integrate from x1 to x2.
    Sorry for bad english.
     
  8. Jan 20, 2005 #7
    yea point A is (6,-3), sorry about that typo. ok i think i see the two triangles and the rectange now. So what i did was found the total area of the two triangle above the x-axis using 1/2*b*h. after that, i found the area of the rectange under the x-axis using L*W. subtract the area of above x-axis with the below and found an answer of 21 which is incorrect. can someone show me what i'm doing incorrect? i think i'm doing the calculations correctly (double checked acouple of times).
     
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