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Homework Help: Find an equation for the tangent line

  1. Apr 6, 2010 #1
    suppose the function F is define by F(x)=[integral from 1 to √x] (2t-1)/(t+2)dt for all real numbers x≥0

    A.) evaluate F(1)
    B.) Evaluate F'(1)
    C.)Find an equation for the tangent line to the graph of F at the point where x=1
    D.) on what intervals is the function F increasing? justify your answer


    please help if you understand, im sorry about the integral sign i dont think there is a symbol for that.
     
  2. jcsd
  3. Apr 6, 2010 #2
    Re: Integration

    You have

    [tex]F(x) = \int_1^{\sqrt{x}} \frac{2t-1}{t+2}\ dt[/tex]

    (A) Write down exactly what F(1) is using your formula and stare at the limits of integration for a few seconds.
    (B) Consider the Fundamental Theorem of Calculus (you will need to apply the chain rule, too)
    (C) Should be easy once you have (A) and (B)
    (D) How does the sign of the derivative of any function relate to whether the function is increasing/decreasing?
     
  4. Apr 6, 2010 #3
    Re: Integration

    i know it should be simple, i dont know how to integrate 2t-1/t+2 help me out
     
  5. Apr 6, 2010 #4
    Re: Integration

    Try substituting u=t+2.
     
  6. Apr 6, 2010 #5
    Re: Integration

    ok what about part b i used the fundamental rule but what about finding it?
     
  7. Apr 6, 2010 #6

    Mark44

    Staff: Mentor

    Re: Integration

    Alternatively, you can just divide 2t -1 by t + 2 using polynomial long division.
     
  8. Apr 6, 2010 #7

    Mark44

    Staff: Mentor

    Re: Integration

    Show us what you did in using the Fundamental Theorem of Calculus.
     
  9. Apr 7, 2010 #8
    Re: Integration

    Don't focus on what the integrand is; focus on the limits. Let's just call the integrand g(t). When x=1, you get

    [tex]F(1) = \int_1^{1} g(t)\ dt[/tex]

    Again, stare at the limits of integration... can you visualize the "interval" over which you are integrating?

    The point is that while you _can_ integrate, you actually do not need to do ANY sort of computation if you take a closer look at your integral.
     
    Last edited: Apr 7, 2010
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