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One integral problem + a couple of others problems.

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Okay first one looks like this:

    Solve
    1
    ∫ (x2+x)dx
    0

    2nd: A bunny set the world record in 1997 at jumping the highest of all the other bunnies. The bunny's jump under the competition looks like: h(x)=4x-4x2. Where "h" is the height in meters and x is the distance from where the bunny started jumping.

    Use the derivitive to calculate the bunny's highest jump.

    3rd: Determine the coordinates of the extreme points of the curve y=x3-3x+5

    3. The attempt at a solution

    1st: Never worked with this before in my life and I understand nothing about this, my textbook gives me an example like this :
    4------------4
    ∫x3dx = [x4/4] = 64-4 = 60.
    2------------2
    This example should be relevant to my problem as it's fairly similar but I don't understand how they got to x4/4 in the example.


    2nd: I tried solving using the derivitive like this: h(x)=4x-4x2. h'(x)=4-8x and then putting -8x on the other side so 8x=4 x=2 but that fails since the answer is supposed to be 1m.

    3rd:
    y=x3-3x+5
    y'=3x2-3=0

    x2 -1=0 Now what? am I on the right track?
     
  2. jcsd
  3. Nov 26, 2013 #2
    1) You really need to be able to understand that textbook example before you can do the question. Maybe review your integration notes and look at it again once you know how to do those general integrals. As you may recall, you can distribute the integral sign to both terms. It's actually just two simple power integrals.

    2) You pretty much had it...until you divided 4 by 8 to get 2 -_- x = 1/2.

    3) You're just about done. You would have two extrema, but the question asks for the coordinates, so you'd want to plug your x coordinates back into your initial equation to get the corresponding y coordinates.
     
  4. Nov 26, 2013 #3

    About 1) I don't have any notes about integrals, I'm just starting with this. I have absolutely no idea how this works.

    2) I didn't divide 4 by 8? 8x=4 x=2. Answer is supposed to be 1m so I don't know where I went wrong.

    3) How do I get out the 2 extrema? x= +- √-1 so x1 = -1 and x2 = 1? And how do I go about plugging this in?
     
  5. Nov 26, 2013 #4

    Mark44

    Staff: Mentor

    Problems involving integration or differentiation definitely should not go into the precalculus section. That's why I moved this thread to the calculus section.
    Integration involves finding the antiderivative. In the example, (1/4)x4 is an antiderivative of x3 because going the other way, d/dx((1/4)x4) = x3. For your first problem, you need to find a function such that d/dx(??) = x2 + x.
    You didn't divide correctly. 4/8 ≠ 2
    Plug these x values into the formula for your function.
     
  6. Nov 26, 2013 #5
    3 and 7 so is the answer 1.3 and 1.7?
     
  7. Nov 26, 2013 #6

    Ray Vickson

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    2nd) Do you really believe that 4/8 = 2? Did the question ask for a value of x?

    3rd) How do you go from x^2 = 1 to x = ± √(-1)?
     
  8. Nov 26, 2013 #7

    vela

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    For problem 2, reread the problem statement and note what are you being asked to find. Don't just assume because you "solved for x" that you solved the problem.
     
  9. Nov 26, 2013 #8

    vela

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    Where did 1.3 and 1.7 come from?
     
  10. Nov 26, 2013 #9
    In OP I showed my calculation at problem 3 i.e this specific problem that you're reffering to, and all I had to do next was this: x2-1=0, x = +-√1 so that's where 1 and -1 so that's the 2 extrema points. and I plugged in those values in the equation y=x3-3x+5. I miswrote as it's obviously supposed to be 1.3 and -1.7.
     
  11. Nov 26, 2013 #10

    3rd, where did I write x^2=1? I wrote x^2-1=0 and from there I used this formula:

    x^2+px+q=0 --> x=-p/2 +-√(p/2)^2-q (the sqrt is covering the entire last part.)

    2nd, I don't know how to figure this out honestly
     
  12. Nov 26, 2013 #11

    vela

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    You mean (1,3) and (-1,7). I'd say you need to pay much more attention to details. You shouldn't be relying on others to figure out what you really mean based on what you say. When I see 1.3 and -1.7, I see two numbers; I don't think of two points, which is what you intended. Similarly, above you wrote ##x = \pm\sqrt{-1}##. That negative sign inside the radical should have immediately told you something went wrong, but you seemed to simply ignore it because, I'm guessing, you thought the sign was an unimportant detail. Finally, in the first problem, you had 8x=4, and you ended up calculating x=8/4 instead of x=4/8 and got the wrong answer. You knew to divide one number by the other, but you didn't bother to make sure you got the order correct.
     
  13. Nov 26, 2013 #12

    Now you are making assumptions about me without knowing me and I don't appreciate it.

    How I got to 1.3 and -1.7 is all there in the thread. Yes I didn't bother to write the whole thing out over and over again in different posts as I had already written it in this thread where it's easy to read the above text. To those whom I responded to regarding problem 3 I wrote it in a way that is easy for them to understand given the information in their response to this problem and that's kind of the whole point in responding to another person, making sure that they'll understand your response. Yes the "-1" in "√" is a mistake I made, it's supposed to be a positive 1 there.

    As for "You knew to divide one number by the other, but you didn't bother to make sure you got the order correct.". I do bother making sure which one to get correct, but if I simply don't understand how to go about this problem it's hard for me to know how to solve it, correct?
     
  14. Nov 26, 2013 #13
    That you're supposed to be writing (1.3) & (-1.7) I didn't know when you're talking about points. That makes a lot of sense though so I apologize about that.
     
  15. Nov 26, 2013 #14

    Mark44

    Staff: Mentor

    IMO, vela is making reasonable assumptions about you that are based on your work, and not taking the time to look over what your wrote to see if it was reasonable. Writing √(-1) and x = 2 in the equation 8x = 4 are examples of this.
    (1) No it isn't, and (2) these don't look like points - they look like two numbers that are completely unrelated to this problem. If you had written (1, 3) and (-1, 7), we might have gotten the idea that you were listing two points. How are we supposed to know that 1.3 means (1, 3)?


    What you wrote for problem 3 was NOT easy to understand. See my comment above.
    That mistake had nothing to do with understanding how to go about that problem. You had the right overall approach to solving the problem, but where you made your mistake was in solving a very simple algebra equation; namely, 8x = 4.

    We all make mistakes, but what you could have done is check your solution. Does 8(2) = 4? Obviously not, so that should have been a clue if you had taken the time to check your answer.
     
  16. Nov 26, 2013 #15


    You said "Plug these x values into the formula for your function." Seemed like you understood it perfectly well, maybe I should've written the formula in the solution to make it easier to understand how I got x^2-1=0 to x=+-√1. I'm sorry you guys, I really do appreciate all your help and I will try my hardest to be more clear for future responses.
     
  17. Nov 26, 2013 #16
    Oh my god I am the slowest person on this planet, of course 8x=4 isn't 2. Okay so that is 4/8=0.5.

    So at problem 2 after I've used the derivitve to get 8x=4, x = 0.5 I plug that value in h(x) = 4x-4x^2 and get 4*0.5-4*(0.5*0.5) which is 1. So the bunnys highest jump is 1m. How can I possibly be this retarded and not see it right away.
     
  18. Nov 26, 2013 #17

    vela

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    I wasn't trying to offend you, but hoping to offer some constructive criticism. Yes, I am making assumptions about you. We invariably make assumptions about students here based on the class they're taking, their work, etc. People who are trying to learn calculus, as you appear to be, typically have a decent background in algebra and trig, and for them, being able to solve 8x=4 for x, realizing that ##\sqrt{-1}## is not correct in this context, and knowing the notation for specifying a point are basic skills. Given this, the mistakes you've made appear to be largely avoidable if you had simply paid more attention to details.
     
  19. Nov 26, 2013 #18

    I understand, no I am not trying to learn calculus. Well guess a part of it however integrals is only a very small part of this course as most of it contains derivitives and graphs. I do need to pay more attention to details, you are right. Looking over my work I missed some obvious things. I'm sorry.
     
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