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Prove value of definite integral is greater than b

  1. May 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove ##\int_{0}^{b} \frac{e^x}{1+x} dx>b## for every ##b>0##.

    3. The attempt at a solution
    Honestly, no idea. I suppose I cannot just calculate the definite integral and estimate. I think it needs to be proven in a more rigorious way.
     
    Last edited: May 9, 2016
  2. jcsd
  3. May 9, 2016 #2

    blue_leaf77

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    Consider the fact that ##\int_0^b 1 dx = b##.
     
  4. May 9, 2016 #3
    Hmm... if ##f(x)>g(x)##, then it follows that ##\int_{a}^{b} f(x) dx## > ##\int_{a}^{b} g(x)dx##?

    I need to show that ## \frac{e^x}{1+x}>1##?
     
  5. May 9, 2016 #4

    blue_leaf77

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    Yes, but only within the interval ##[0,b]##.
     
  6. May 9, 2016 #5
    The first hint you gave was very helpful.

    Let ##0≤x≤b##,

    then ## \frac{e^x}{1+x}≥1##⇔## e^x≥1+x## which is true since we know the expansion ##e^x=1+x+\frac{x^2}{2}+...≥1+x##

    Therefore ##\int_{0}^{b} \frac{e^x}{1+x} dx## ≥ ##\int_{0}^{b} 1 dx=b##

    and ##\int_{0}^{b} \frac{e^x}{1+x} dx## = ##\int_{0}^{b} 1 dx=b## iff ## x=0##.

    I think I am almost there...is there a mistake in the problem statement because ##\int_{0}^{b} \frac{e^x}{1+x} dx## can have value of b when ##x=0##? I think it wouldn't contradict anything, would it?

    if ##x=0##, then ##f(x)=\frac{e^x}{1+x}## ruduces to constant function so maybe that's why we assume ##x≠0##?
     
    Last edited: May 9, 2016
  7. May 9, 2016 #6

    Ray Vickson

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    No. When ##x = 0## we have ##f(x) = 1##, but ##f## is NOT the constant function!

    Anyway, on ##[0,\infty)## you have ##f(x) \geq 1##, with strict inequality when ##x > 0##.
     
  8. May 9, 2016 #7

    blue_leaf77

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    Thanks to Ray I know that the OP edited his last post after the last time I checked it.
    This line is unnecessary. Moreover those integrals are not a function of ##x## (if you want you can define it as a function of ##b##, though), thus evaluating it at certain value of ##x## makes no sense. It's like you want to evaluate a function ##f(y)## at ##x=0##.
    $$
    \int_{0}^{b} \frac{e^x}{1+x} dx = b
    $$
    is true if when ##b=0##, but the problem states that the condition for ##b## is that it is bigger than zero, so there is no contradiction.
     
  9. May 11, 2016 #8
    Is that obvious or need to be proven?
     
  10. May 11, 2016 #9

    blue_leaf77

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    It's obvious but why do you ask about it now?
     
  11. May 11, 2016 #10
    I am asking about it now because I am checking and reviewing my work. This piece of homework is due to tomorrow. It is very obvious to me that ##\int_0^b 1 dx = b##, but we haven't proven that in class and I'm wondering if I am supposed to prove that as well or if I can just say to my teacher that "oh, that's obvious." And it's better ask late than never, right?
     
    Last edited: May 11, 2016
  12. May 11, 2016 #11

    blue_leaf77

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    Then in that case, prove it. It should require only one line.
     
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