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Homework Help: Definite integral problem

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    If the value of the integral ##\displaystyle \int_1^2 e^{x^2}\,\, dx## is ##\alpha##, then the value of ##\displaystyle \int_e^{e^4} \sqrt{\ln x} \,\, dx## is:


    2. Relevant equations

    3. The attempt at a solution
    Starting with the given integral, I used the substitution, ##e^{x^2}=t\Rightarrow 2xe^{x^2}dx=dt##.
    $$\int_1^2 e^{x^2} dx=\int_1^2 \frac{2xe^{x^2}}{2x}dx=\frac{1}{2}\int_e^{e^4} \frac{dt}{\sqrt{\ln t}}$$
    But this doesn't end up with the definite integral asked in the problem. :(

    I have tried using the substitution ##\sqrt{\ln x}=t## in the definite integral to be evaluated, I end up with ##\displaystyle \int_1^2 t^2\cdot e^{t^2} dt## but this isn't the same as given in the problem statement.

    Any help is appreciated. Thanks!
  2. jcsd
  3. Oct 15, 2013 #2


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    Hi Pranav-Arora! :smile:

    Try going the other way …

    start with ∫ √(lnt) dt, and make the same substitution. :wink:
  4. Oct 15, 2013 #3
    I have already tried substituting ##\sqrt{\ln x}=t##. Do you ask me to use the substitution ##x=e^{t^2}## in the definite integral to be evaluated?
  5. Oct 15, 2013 #4


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    True, but you can then integrate by parts.
  6. Oct 15, 2013 #5
    How? :confused:

    Integrating by parts:
    $$t^2\int e^{t^2}dt-\int \left(2t \int e^{t^2}dt \right) dt$$

    How do I proceed further?
  7. Oct 15, 2013 #6


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    t2et2 = t(tet2) :wink:
  8. Oct 16, 2013 #7
    This is an extension of an earlier problem you posted involving differentiation under the integral sign using Leibnitz Rule. It does not require integration by parts.

    Let ##I=\displaystyle \int_1^2 e^{x^2}\,\, dx##

    Substitute [itex]x = \sqrt{β}t[/itex] to obtain:
    [tex]I=I(β)=\sqrt{β} \int_{\frac{1}{\sqrt{β}}}^{\frac{2}{\sqrt{β}}} e^{βt^2}\,\, dt[/tex]

    As in the earlier problem, the next step is to take the derivative of I with respect to β and evaluate this derivative at β=1. This requires use of the Leibnitz Rule not only for the function under the integral sign, but for the limits of integration as well. Since, in reality, the derivative of I with respect to β must be zero (since the original form of the integral does not involve β), you will end up with the sum of four terms equal to zero. You can then solve for the integral you really want.

    It only took me a couple of minutes to solve the problem this way (most of which involved making sure that I did the differentiation algebra correctly). My final answer was one of the four choices given.

    Last edited: Oct 16, 2013
  9. Oct 16, 2013 #8
    Great! Thanks a lot tiny-tim! :)

    Can you please post the link to the thread you are talking about? I don't think I posted a similar problem before.
    Okay, I tried this and I was able to evaluate ##\displaystyle \int_1^2 e^{x^2}\,\, dx## but the problem doesn't ask this, how do I go about evaluating what's asked? Do I have to use the same substitution?
    But tiny-tim's method is nice too. :)

    Anyways, in my syllabus, use of differentiation under the integral sign is limited. I have done problems where the function to be integrated is not a function of the same variable as of limits, I haven't yet encountered a problem where this is not the case.
  10. Oct 16, 2013 #9
    When I differentiate the function I(β) with respect to β, I get:
    [tex]\frac{dI}{dβ}=\frac{1}{2\sqrt{β}} \int_{\frac{1}{\sqrt{β}}}^{\frac{2}{\sqrt{β}}} e^{βt^2}\,\, dt+β^{\frac{3}{2}}\int_{\frac{1}{\sqrt{β}}}^{\frac{2}{\sqrt{β}}} t^2e^{βt^2}\,\, dt+\frac{1}{β}\left(-e^4+\frac{e}{2}\right)[/tex]
    This expression must be equal to zero; and, if I set β equal to 1, I get:
    [tex]\frac{1}{2} \int_{1}^{2} e^{t^2}\,\, dt+\int_{1}^{2} t^2e^{t^2}\,\, dt+\left(-e^4+\frac{e}{2}\right)=0[/tex]
    Rearranging this equation and multiplying it by 2 yields:
    [tex]2\int_{1}^{2} t^2e^{t^2}\,\, dt=2e^4-e-\int_{1}^{2} e^{t^2}\,\, dt=2e^4-e-α[/tex]
    This is answer B.
  11. Oct 16, 2013 #10
    Very sorry, I should have been careful. :redface:

    Thanks a lot Chet! :smile:
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