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Help me proove absolute value inequalities and Solve an Integral

  1. Sep 4, 2014 #1

    RJLiberator

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    1) True or Fale?
    |a-b| [itex]\geq[/itex] ||a|-|b||

    My solution: I broke this up into 4 different cases. 1. a> 0 b>0 2. a<0 b<0 and so on...
    For each case, I ended up with a more simplified version of the inequality. For instance, in case 1 where I used a<0 and b<0, I ended up with the simplified statement of:
    [itex]|-a+b|\geq|-a+b|[/itex] Which is clearly equivalent.

    Is this a correct way to prove that this statement is true?

    2. With that being out of the way, how can I prove this statement to be true:
    [itex]\sqrt{(a+b)^2} = |a+b|[/itex]
    Would I do the same thing and break it up into 4 cases?

    3. This has been my most difficult problem so far.

    For any x > 0, consider the function

    f(x)=[itex]\int^{\sqrt{x}}_{0}e^{t^{2}}dt[/itex]

    Computer f'(x)

    My work so far: I was told to use the chain rule where the inner function is [itex]\sqrt{x}[/itex] and the outer is 0.

    I'm not qutie sure what to do.. I imagine I take the antiderivative of this function using the chain rule?
     
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2

    HallsofIvy

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    If you are only considering "1. a> 0, b> 0 2. a< 0 b< 0 and so on" HOW did you get to [itex]|-a+ b|[/itex]? You also need to consider, in each of those cases a< b and a> b.

    You know that |a|= |b| if and only if [itex]a^2= b^2[/itex], don't you?

    This makes no sense. The left side is a function but the right side is a differential. they can't be equal. Did you mean [itex]f(x)= \int e^t dt[/itex]? If so finding f' is just the "fundamental theorem of Calculus".

    ??? There is NO square root in the problem you gave- and an "outer function" of 0 would just make the entire function 0.

    You said the problem was to find the derivative not the anti-derivative. What, exactly is the problem.
     
  4. Sep 4, 2014 #3

    RJLiberator

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    Halls, Thanks for the reply:

    Regarding 1)
    To save space I cut my claims short. I did take into account a>b and b>a

    2) So if I square both sides, I would receive (a+b)^2 = (a+b)^2 Hm. That makes a lot of sense.

    3) It seems you may have quoted my question when I was editing/had a problem with the latex. The proper latex version is up now.
    With that being said, I understand why the outer function of 0 would make the chain rule null. Maybe my TA was not looking at the problem carefully enough.

    I will need to search through the Fundamental theorem of Calculus then.

    Thank you for the input. It is appreciated.
     
  5. Sep 4, 2014 #4

    LCKurtz

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    Since you apparently solved it using many cases, I will suggest an easier way not needing all those cases. Start with$$
    |a| = |(a-b)+b| \le |a-b| + |b|$$(I'm assuming you already have the triangle inequality here). So this gives you$$
    |a|-|b| \le |a-b|$$. Now do you see how this same result with ##a## and ##b## switched gives you your required inequality?
     
  6. Sep 4, 2014 #5

    RJLiberator

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    Hey guys, I think I figured out the third problem. The integral problem using the fundamental theorem of calculus. Can anyone check my work:

    1. denote a new function :
    g'(x) = e^(x^2)
    f(x)=g(sqrt(x))

    Using the chain rule
    f'(x) = 1/2(sqrt(x) * e^(sqrt(x)^2)

    How does this look?




    LCK, I am going to look into what you posted. One moment.
     
  7. Sep 4, 2014 #6

    HallsofIvy

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    Surely, you can simplify "(sqrt(x))^2".
     
  8. Sep 4, 2014 #7

    RJLiberator

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    Excellent. e^x.

    =)
     
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