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Integrating 2nd order derivative

  1. Aug 22, 2009 #1
    The question is

    If f(x) = 7x^3 + 8x^2 - x + 11, evaluate :

    a, Integral +1 - -1 f(x) dx
    b, Integral +1 - -1 f'(x) dx
    c, Integral +1 - -1 f''(x) dx

    For a, Just integrate each individual and then input the figures which gave me

    1.75x^4 + (8x^3)/3 - 0.5x^2 + 11x

    Which when I input the figures gives me 27 1/3.

    It is b, which I am unsure about. Do I intergrate 1.75x^4 + (8x^3)/3 - 0.5x^2 + 11x and then put the values in?

    Some guidence would be most appreciated, thank you.
  2. jcsd
  3. Aug 22, 2009 #2

    Gib Z

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    Homework Helper

    Hello Maths Muppet! Welcome to PF.

    Are you familiar with the fundamental theorem of Calculus? The form of it that is useful here is:

    [tex]\int^b_a g'(x) dx = g(b) - g(a)[/tex]

    You can apply that to b and c quite directly.
  4. Aug 24, 2009 #3
    I think your reply has just confused me a little bit. I might be using the wrong termonology but I thought all that I would have to is integrate 1.75x^4 + 8x^3 - 0.5x^2 + 11x and then insert the values back in. Is this correct?
  5. Aug 24, 2009 #4


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    f'(x) means the derivative of f(x) with respect to x. Integration is the reverse of differentiation. So for example, if f(x)=x2, then f'(x)=2x. So ∫2x dx=x2+C.

    See now why Gib Z said you can directly work out the integral?
  6. Aug 25, 2009 #5


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    Science Advisor

    1.75x^4+ 8x^3- 0.5 x^2+ 11x is the integral of your original function and does not have to be integrated again for problem (a).

    Gib_z's point is that
    [tex]\int_{-1}^1 f'(x)dx= f(1)- f(-1)[/tex]
    and that
    [tex]\int_{-1}^1 f"(x)dx= f'(1)- f'(-1)[/tex]
  7. Aug 28, 2009 #6
    [itex]\int_a^b f(x)\,dx\,=\,F(b)\,-\,F(a),\,where\,F'(x)\,=\,f(x).[/itex]
    FTOC is great!!!
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