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Homework Help: Find the primitives of the functions secx tanx

  1. Nov 20, 2006 #1
    hey just wondering if someone could help me with these 2 questions!

    Suppose for a function f, that 1<= f'(x)<=3 show that

    I dont even no where to start... I was thinkin that in some way you could use the mean value theorem!
    f(c)= [f(6)-f(2)]/4

    But I wouldnt have a clue where to go from there

    2. Find the primitives of the functions

    so I sed that u=secx and that du=secxtanx
    So does this mean that the answer will just be equal to one?

    The last one is just hard in the fact that I am unsure of wat u should be equal to...

    integral x^6lnx(dx)
  2. jcsd
  3. Nov 20, 2006 #2
    This problem is actually relatively simple, you're probably just overanalyzing it.

    First, let's assume f(2) = 0. Since 1<=f'(x)<=3 for all x then the maximum change over the interval 2 to 6 is M*dx where M is the maximum value of the derivative of the function and dx is the change in x, in this case (6-2) or 4. So, the maximum difference between f(6) and f(2) is 12. If we analyze the minimum difference in the same way using the minimum value of the derivative, we arrive at 4. Therefore, the difference in the values of f(2) and f(6) must be between 4 and 12.

    The second one is also simple, secx*tanx just happens to be the derivative of secx.

    The last one is a simple integration by parts. Let x^6 = dv and ln(x)dx = u.

    ==> Integral(x^6*ln(x) dx) = uv - Integral(v*du) -- Integral(x^6*ln(x) dx) = ln(x)*(x^7)/7 - Integral(x^6/7 dx)

    Or -- ln(x)*x^7/7 - x^7/49
    Last edited: Nov 20, 2006
  4. Nov 20, 2006 #3
    2. [tex] \int \sec x \tan x \; dx = \sec x + C [/tex]
  5. Nov 20, 2006 #4
    yeah I kinda get that... little bit confusing in some ways... y is it that u can just assume that f(2)=0... I get how u get the answers 4 and 12 from there then!
  6. Nov 20, 2006 #5
    qu 2 yeah whoops thats wat I meant... I dont know y I sed it was equal to one!
  7. Nov 20, 2006 #6
    Well it doesn't matter what initial value we choose for f(2) since the change of the function over a given interval is bounded by the definition of the derivative given in the problem. I could have chosen 9,042 and I still would have gotten f(6) at max is 9,054 and at minimum 9,046. Reread my original post very carefully and think about how differentials work and it should become pretty clear what I'm saying.
    Last edited: Nov 20, 2006
  8. Nov 20, 2006 #7
    cool thanks... so its just easier to choose f(2)=0!
    Thanks agen!
  9. Nov 20, 2006 #8
    any suggestions for part 3... I know u need to integrate by parts... any pointers would be greatly appreciated
  10. Nov 20, 2006 #9
    Taryn, for your last problem:

    [tex] \int x^6 \ln x dx [/tex]

    Let [tex]u = \ln x[/tex] and let [tex]dv = x^6 dx[/tex]. It follows from there that: [tex]du = \frac{dx}{x}[/tex] and that [tex]v = \frac{x^7}{7}[/tex].

    It's all integration by parts from there:

    [tex] \int u dv = uv - \int v du[/tex]
  11. Nov 21, 2006 #10


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    For the first problem just integrate all 3 terms of the initial double inequality between 2 and 6.

  12. Nov 21, 2006 #11
    yeah okay wat I also did was letting u=x^7, du=7x^6 which is du/7=x^6
    Then I got int 1/7(lnu^1/7)du
    Then I used log laws and so I then had 1/49 int (lnu)(du)
    I ended up gettin the answer (x^7lnx)/7 - (x^7)/49 + C
    Is this rite?
  13. Nov 21, 2006 #12
    Thats cool anyhow I know I did... checked it using your way! Thanks for your time!
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