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Definite integrals

  1. Oct 4, 2003 #1
    I'm now teaching myself several topics on definite integrals for a math test on monday. Here are a few problems that I don't know how to do.

    Q1) Prove the following inequality:
    1 < [inte]pi/20 (sin x)/x dx < (pi/2)

    Q2) Show that for x > 0,
    ex-1 <= [inte]x0 (e2t+1)1/2 dt <= 21/2(ex-1)

    Q3) Let F(x) = [inte]2x0 et/x dt. Find F'(x)

    Any help would be appreciated.

    Edit: I know how to do Q3 now but I still don't understand why the upper limit can change to 2 after substituting u = t/x
    Last edited: Oct 4, 2003
  2. jcsd
  3. Oct 4, 2003 #2
    Q4) Let f be an odd function, show that
    &int;-a...a(&int;-a...xf(t)dt)dx = - &int;-a..a xf(x)dx
  4. Oct 4, 2003 #3


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    This is due to your change in variables, you must also do the change of variables to the limits. t=2x is your upper limit, your change of variables is u=t/x -> u = 2 if t=2x.
  5. Oct 4, 2003 #4


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    Try doing Q1) by replacing (sin x)/x with its taylor series. This will give both inequalities... however the right inequality is easy to get by considering the max value of (sin x)/x

    (this feels awfully brute-forcish, though... there's probably a slicker way)

    For Q2), consider the fact that the inequalities hold when x = 0, then consider how fast each term grows as x increases.

    Another approach to Q3) is to recognize that you can simply do the integral directly.

    For Q4), have you tried switching the order of integration?
  6. Oct 4, 2003 #5
    ok, I got it now. Thanks.

    Yep, I can do the right inequality now. I haven't learnt taylor seies in school yet, so I still can't get the left one.

    Could you please show me the first few steps? Q2 was extracted from the section about inequalities on definite integrals and mean value theorem of integral calculus. After reading the theorems and examples, I attempted to do questions on this topic. I tried the first 4 questions (the easiest ones) but could only do 2 of them. I really felt frustrated and left this section behind and read the next section. (Edit: I'll go back to this topic after finish revising other topics)

    Could you please show me? I'm new to fundamental theorem of integral calculus.

    I attempted the question as follows:
    g(x) = &int;-a...xf(t)dt ...................... (1)
    g'(x) = f(x)
    integrate on both sides
    g(x) = &int;f(x)dx = xf(x) + C = -xf(x) + C (f is an odd function)

    Now I want C=0
    From (1), g(-a) = 0
    af(-a) + C = 0
    C = af(a)
    Hm.. something must have gone wrong but I don't know where.

    I'm not sure what do you mean by switching the order of integration. It's my first time doing question involving 2 integral signs writing together.
    Last edited: Oct 4, 2003
  7. Oct 4, 2003 #6


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    All right...

    My next best suggestion is to divide the domain of integration into several parts (hint: 4 will work) and find a simple minimum value over each of the 4 parts.

    Again, sorry I can't figure out an elegant solution; I'm just as annoyed about it as you are. :smile:


    I'll do a problem to demonstrate the technique I'm suggesting.

    Theorem: 2x <= 2x for x >= 2

    let f(x) = 2x - 2x
    First, notice that f(2) = 0.
    Now, f'(x) = ln(2) 2x - 2
    Clearly, f'(x) > 0 for x >= 2, so f(x) is strictly increasing over [2, &infin;)
    Thus, 2x - 2x >= 0 for x >= 2
    Therefore 2x <= 2x for x >= 2

    (I'm unsure if this type of proof is the way your text means for you to do the problem, but no other solution has sprung to my mind)


    First let me ask if you know how to compute something like:

    &int;[0,1] e3x dx

    (this means the integral from 0 to 1)

    If you can, then note that, as far as the integral is concerned, (1/x) is a constant, just like 3 is in the above problem.


    The last two steps here are incorrect.

    As for switching the order of integration, forget about it; I presumed you were in a multivariable calc course where you would learn such a thing.

    Try integration by parts... on the outer integral.
  8. Oct 10, 2003 #7
    Thanks Hurkyl. Sorry that I haven't had time to reply immediately.
    Eek, that was an awful mistake.

    I got it now.

    By applying integration by parts,

    =[ x&int;-a..xf(t)dt ](-a..a) - &int;xf(x)dx
    =[ a&int;-a..af(t)dt - (-a)&int;-a..-af(t)dt ]- &int;xf(x)dx

    since f(x) is an odd function, so (a&int;[-a..a]f(t)dt) = 0
    since the upper limit = lower limit, so (-a)&int;[-a..-a]f(t)dt = 0

    The result follows.

    Yes, I know &int;[0,1] e3x dx = (1/3)&int;[0,1]e3xd(3x) = (1/3)(e^(3x)-1)

    Applying this method,

    = &int;[0..2x] et/x dt
    = -(x2/t)&int;[0..2x]et/xd(t/x)

    If I need to find F'(x), I have to apply chain rule and isn't a nice job to continue from here.

    It seems to me that method is simliar to do substitution, or am I misunderstand something?

    I'm still working on Q1 and Q2. I've been taught the technique that you used as demonstration for Q2 but I couldn't think of it while attempting this question. :wink:
  9. Oct 10, 2003 #8


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    In general...

    &int; eat dt = (1/a) eat + C

    so if we plug in 1/x for a, we get...

    &int; et/x dt = x et/x + C

    Then you can evaluate at the endpoints of [0, 2x] to get the exact answer, at which point differentiation is easy.
  10. Oct 10, 2003 #9
    Oh, we can evaluate the integral first before differentiating it! I see. :smile:
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