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Homework Help: Apply the mean value theorem for integrals

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    My book is not explaining very well the steps at solving these problems. There is a step that I'm missing

    step 1. find 1/(b-a) easy
    step 2. find the antiderivative of 4 - x, easy, x^2/2
    step 3. plug in what the result of the antiderivative is for b and a, subtract b from a
    step 4. this is the step I'm missing. the books shows that I'm supposed to multiply 4 by b - a. why? what is this? I need to know what mathematical terminology this is so that i can do these problems in the future
    step 5. subtract step 4 from step 3. why? what mathematical terminology is this.
    step 6. multiply step 1 by step 5, I understand that.

    I'm not yet able to read the notation and the book is not helping.
  2. jcsd
  3. Feb 1, 2012 #2


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    Science Advisor

    Well, 4x- x^2/2

    The anti-derivative of 4, with respect to x, is 4x. That is evaluated between the limits of integration, b and a. You can do it as either 4b- 4a or 4(b- a). They are equal by the "distributive law".

    No, they are not subtracting step 4 from step 3. They are simply completing the calculation indicated in step 4. There are, however, different ways of doing that step. You could have calculated that 4(3)= 12, and [itex]3^2/2= 9/2[/itex] and subtract: 12- 9/2= 24/2- 9/2= 15/2.
    Then multiply by the leading 1/3: (1/3)(15/2)= 5/2.

    Or, using the distributive law, go ahead and multiply by the 1/3 first: (1/3)(4(3)- 3^2/2)= (1/3)(4(3))- (1/3)(3^2)/2. That 1/3 will cancel a 3 in eacy term: it is equal to 4- 3/2= 8/2- 3/2= 5/2 again.

    The distributive law: a(b+ c)= ab+ ac.
  4. Feb 1, 2012 #3
    Ok, thanks, I incorrectly calculated the antiderivative of 4 - x as (x^2)/2, it's (4x - x^2)/2.

    Thanks for your help
  5. Feb 1, 2012 #4
    All that's happened here is an integral
    Here are some properties of integrals that you should know that will also help you understand your problem

    Let a be a constant, f(x) be some function of x, and F(x) be it's anti-derivative

    1. [itex]\int (a \pm f(x))dx = \int a dx \pm \int f(x) dx[/itex]

    2. [itex]\int a f(x) dx = a \int f(x) dx[/itex]

    3. [itex]\int 1 dx = x[/itex]

    4. [itex]\int_a ^b f(x) dx = F(b) - F(a)[/itex]

    Now I shall show you which of these properties we shall use in our problem;

    [itex]\frac{1}{b-a} \int_a ^b f(x) fx = \frac{1}{b-a} \int_a ^b (4-x)dx [/itex]

    I shall use property 1. now

    [itex]\frac{1}{b-a} \int_a ^b (4-x)dx [/itex] = \frac{1}{b-a}( \int_a ^b 4 dx - \int_a ^b x dx[/itex]

    I shall now use property 2. on the first term

    [itex] \frac{1}{b-a}( \int_a ^b 4 dx - \int_a ^b x dx = \frac{1}{b-a} ( 4 \int _a ^b 1 dx - \int_a ^b x dx)[/itex]

    Due to property 3, the anti-derivative of 1 is x (or in the terms of property 4. if f(x)=1 then F(x) = x)
    And using property 4 to evaluate the definite integrals

    [itex]\frac{1}{b-a} ( 4 \int _a ^b 1 dx - \int_a ^b x dx) = \frac{1}{b-a} ( 4(b-a) -( \frac{b^2}{2} -\frac{a^2}{2} ))[/itex]

    Substituting in our values for a and b we get

    [itex] \frac{1}{b-a} ( 4(b-a) -( \frac{b^2}{2} -\frac{a^2}{2} )) = \frac{1}{3} ( 4(3) -\frac{3^2}{2}) = 4 - \frac{3}{2} = \frac{5}{2}[/itex]

    Has this helped at all?
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