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Integration (2)

  1. Jan 22, 2007 #1
    Theorem: Let f be continuous on [a,b]. The function g defined on [a,b] by [​IMG]
    is continuous on [a,b], differentiable on (a,b), and has derivative g'(x)=f(x) for all x in (a,b)


    1) Given that g is defined by g(x)=definite integral of sin (pi t) from 0 to x for all real numbers x, determine g'(-2), if it exists.
    [In this case a=0, according to the above theorem g'(x)=f(x) for all x in (a,b), but the theorem doesn't say that g'(x)=f(x) for x<a, and in this case -2<0=a. Does it mean that g'(-2) does not exist? ]


    2) Let g be defined by g(x)=definite integral of 1/ (1+t^2) from 0 to x for all real numbers x.
    2a) Show that if x<0, then g(x)<0 using properties of integrals.

    [Sorry, I really have no clue about part a and I need some help on it...]
    2b) Find the intervals on which g is increasing or decreasing.
    [According to the theorem on the top g'(x)=1/ (1+x^2) is true only for all x in (a,b), in this case a=0, then how can I find g'(x) for the part x<0 ?]


    I hope someone would be nice enough to help me out! Thanks a lot!
     
    Last edited: Jan 23, 2007
  2. jcsd
  3. Jan 22, 2007 #2
    2) Let g be defined by g(x)=definite integral of 1/ (1+t^2) from 0 to x for all real numbers x.
    2a) Show that if x<0, then g(x)<0 using properties of integrals.

    [Sorry, I really have no clue about part a and I need some help on it...]

    [itex]\frac{1}{1+t^2}[/itex] is positive for all real t.

    Maybe you copied the question down wrong?
     
  4. Jan 22, 2007 #3
    No, the actual function g is
    g(x)=definite integral of 1/ (1+t^2) from 0 to x
    Something like [​IMG]
     
  5. Jan 22, 2007 #4

    mjsd

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    Homework Helper

    I hope I have understood you questions... and can answer it in the spirit they were intended.

    ok, firstly forget about the -2, you want g'(x) evaluated at -2, so you need g'(x) first and foremost. now the theorem tells you that you can actually do this differentiation IF function f(x) is continous on [0,x] (in this case) where x can be any real number. So, you can do differentiation of g(x) and it is just f(x). now is f(-2) defined? and what it is? By the way, no one demands that x got to be bigger than 0....



    using properties of integrals? can't you just integrate and see that the function is always negative? I think this is what it may mean:
    [tex]\displaymath{g(x) =\int_0^x f(t) dt = -\int_x^0 f(t) dt}[/tex]
    now f(t) is always +ve and limits are now going from smaller to bigger number (for x<0 is given) so result of integral itself is always +ve, and then the -ve sign out front ensures that g(x) is always -ve.
    :confused:

    again x can be any real number so can be -ve too... you sure you want f(x) and not g(x)? anyway, function f(x) is well-defined and information may be easily extracted by doing a derivative test. For g(x) though I assume you need to use properties of integrals again...

    after thought: this is not a difficult problem as such, it is just an exercise of justifying all your steps carefully using the theorems you know. There may be other properties or theorem you are meant to use to construct your arguments.
     
  6. Jan 24, 2007 #5
    Hi

    1) By the theorem on the top, I know that g'(x)=f(x) for x>0 is certainly true, because a=0.
    But for x<0, is it still true that g'(x)=f(x)? And how do you know that?

    2a) [tex]\displaymath{g(x) = -\int_x^0 f(t) dt}[/tex]
    I still don't understand the stuff to the right of the integral sign is positive for x<0, could you explain further?

    2b) Yes, there is a typo. The correct question should read "Find the intervals on which g is increasing or decreasing."
    But the same story, according to the theorem on the top g'(x)=1/ (1+x^2) is true only for all x in (a,b), in this case a=0, so g'(x)=1/ (1+x^2) is true for x>0 for sure. How about for the part x<0 or x=0? How can I find g'(x) for the part x<=0 ?]



    Recall: The Theorem says: Let f be continuous on [a,b]. The function g defined on [a,b] by [​IMG]
    is continuous on [a,b], differentiable on (a,b), and has derivative g'(x)=f(x) for all x in (a,b)

    Thanks!
     
    Last edited: Jan 24, 2007
  7. Jan 24, 2007 #6

    mjsd

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    2a) since [tex]f(t)\geq 0[/tex] everywhere, it is quite easily see that
    [tex]\displaymath{\int_a^b f(t) dt \geq 0}[/tex]
    if [tex]b> a[/tex]
    if you want a proof for this, this comes directly from the definition of the definite integral as a limit of Riemann sums (note b>a is essential in this definition for it ensures that [tex]\Delta y_i[/tex] is non-negative)
    [tex]\displaymath{
    \int_a^b f(y) dy = \lim_{\max \Delta y_i\rightarrow 0}
    \sum_{i=0}^n f(y_i^*) \Delta y_i}[/tex]

    here [tex]y_i^*[/tex] is some value in each sub-interval [tex]\left[x_{i-1}, x_i\right][/tex], [tex]\Delta y_i = (b-a)/n[/tex]

    now when f(t) is non-negative, all individuals terms in the sum are non-negative so the integral is non-negative.
     
    Last edited: Jan 24, 2007
  8. Jan 24, 2007 #7

    HallsofIvy

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    You don't have to take a= 0. [itex]sin(\pi t)[/itex] is continuous for all t. For example,
    [tex]\int_{-100}^x f(t)dt= \int_{-100}^0 f(t)dt+ \int_0^x f(t)dt[/itex]
    The first integral on the right is a constant and so has derivative 0. That means that the integral on the left and the second integral on the right have the same derivative. Apply the theorem with a=-100, b= 0. -2 is certainly in that integral.

    [itex]f(t)= \frac{1}{1+ t^2}[/itex] is an even function and positive for all t. Clearly, if x< 0, then [itex]\int_x^0 \frac{1}{1+ t^2}dt[/itex]
    is positive (it is the area between the graph of [itex]y= \frac{1}{1+t^2}[/itex] and the x-axis between x and 0 and area is always positive). But [itex]\int_0^x \frac{1}{1+t^2}dt= -\int_x^0 \frac{1}{1+t^2}dt[/itex]

    The "a" and "b" are red herrings. Changing the "a" in the integral just changes the additive constant which is irrelevant to the derivative. As long as f(x) is continuous at x0, the derivative of g at x0 is f(0). Since [itex]\frac{1}{1+t^2}[/itex] is continuous for all t, [itex]g'(x)= \frac{1}{1+x^2}[/itex] for all x. Of course, a function is increasing as long as its derivative is positive, decreasing if its derivative is negative.


     
  9. Feb 1, 2007 #8
    Many thanks for the responses, I think I am understanding it better now...


    The top theorem says if f is continuous on [a,b] , then g is also continuous on [a,b]

    Let me define a piecewise function
    f(x)=x^2+x, if 0<=x<=1
    f(x)=2x, if 1<x<=3
    Clearly, f is continuous on [0,3]

    Let F(x)=definite integral of f from 0 to x, with 0<=x<=3
    i.e.,
    F(x)=x^3/3 + x^2/2, if 0<=x<=1
    F(x)=x^2, if 1<x<=3
    When I graph F, it has a discontinuity at x=1
    So F is not continuous on [0,3], why is it like that? The theorem clearly states that f continuous implies F continuous, am I right? Or did I do something wrong?
     
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