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Really, really basic integration!

  1. Apr 27, 2007 #1
    Just double checking my answers basically, its been a while since I did this..

    http://img220.imageshack.us/img220/4075/65177751rq6.jpg [Broken]

    My answer is 14

    http://img220.imageshack.us/img220/3291/92105764te4.jpg [Broken]

    My answer is -22

    http://img223.imageshack.us/img223/4874/13030647ds3.jpg [Broken]

    My answer is 23

    http://img223.imageshack.us/img223/5766/95610620ke0.jpg [Broken]

    My answer is 4

    Hope I'm right! :rofl:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 27, 2007 #2


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    Your images don't seem to be loading in my comp, although I don't know if it loads for others. Maybe you can try posting the questions in LaTeX?
  4. Apr 28, 2007 #3
    hope you can see them now! :bugeye:
  5. Apr 28, 2007 #4
    Your last answer is wrong

  6. Apr 28, 2007 #5
    Dammnit, that means I got Q1 wrong too. I just double checked, the area labelled 3 refers to the interval from b to c.
  7. Apr 28, 2007 #6

    Gib Z

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    No t_n_p you are correct!! christianjb didn't look at the bounds close enough!!!!

    However for question 2 we can rewrite as [itex]-2\int^c_a f(x) dx[/itex]. The integral can be seen to be 6-17=-11. Then times by -2. 22, not negative 22.
    Last edited: Apr 28, 2007
  8. Apr 28, 2007 #7
    Are you sure?
    Re Q1, 10-3+10=17?

    Re Q4, 10-3 = 7?
  9. Apr 28, 2007 #8

    Gib Z

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    Well I'm assuming the area labelled 3 in both the diagrams refers to the area between c and the origin, not c and b..
  10. Apr 28, 2007 #9
    Sigh, yeh, 3 refers to interval between c and d.

    So I've got 1/4 right.

    :surprised :yuck: :cry: :grumpy:
  11. Apr 28, 2007 #10

    Gib Z

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    1. 17
    2. 22
    3. 23
    4. 7

    Any confusions just ask.

    And sorry about that christianjb, I guess I jumped the gun.
  12. Apr 28, 2007 #11


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    No one seems to have mentioned problem 1, which seems quite easy, but your given answer, 14, is clearly wrong. In fact, I'm wondering where you could have got that! You are told that the area under the curve on the right is 10. You are also told that the function is even so the curve is symmetric about the y-axis and therefore the area under the curve on the left is also 10. Finally, you are told that the area above that small part in the middle is 3. Because it is below the x-axis the integral itself is negative. The total area is, of course, the sum of the "positive" areas minus the negative "area". And that is not 14!

    Did you perhaps think that the portion, below the x-axis, to the right of the y-axis was 3? That would make the total "negative" area 6 and would give a total of 10+10- 6= 14 for the total area. That is not, however, how I would interpret what is given.
  13. Apr 28, 2007 #12
    Yeah thats what I did in Q1 and 4. I just assumed 3 was from 0 to c, but I've now been told that its from b to c.
  14. Apr 28, 2007 #13


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    The question is WHO told you that? That would be my interpretation but it is your teacher's that is important!:rolleyes:
  15. Apr 28, 2007 #14
    Unfortunately, the teacher told me that...:cry:
  16. Apr 28, 2007 #15
    Aha, I best the mighty Gib Z in integral combat at last! :smile:
  17. Apr 28, 2007 #16


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    Not "unfortunately"! That's exactly who has the right to tell you that!
  18. Apr 29, 2007 #17
    Well it was unfortunate for me...
  19. Apr 29, 2007 #18
    And (unless I'm missing something) no-one seems to have noticed that the limits of the integral in Q1 are from "d" to "a" and not from "a" to "d" as one might expect.

    Hence, the value of the integral is actually -(10 -3 + 10) = -17.
  20. Apr 29, 2007 #19

    Gib Z

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    ::Cries:: Good spot Theo im having a really bad episode..
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