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Process for solving integration problems

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  1. Nov 10, 2015 #1
    . The problem statement, all variables and given/known data
    Q1. Let f(x) be any continuous function that satisfies: $$-2x≤xg(x)≤2x$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

    $$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx$$

    Q2. Let h(x) be any continuous function that satisfies: $$-4≤h(x)≤x^2-4$$ for $$0≤x≤1$$ Find the upper and lower bounds for:

    $$\int_{0}^{1}x^3h(x)dx$$

    3. The attempt at a solution

    Soltn for Q1 for lower bounds: $$\int_{0}^{1} \sqrt{1+g(x)+x^2}dx \\

    = \int_{0}^{1} \sqrt{1-2x+x^2}dx \\

    = \int_{0}^{1} \sqrt{(x-1)^2}dx \\

    L_{f}(P) = \int_{0}^{1} (x-1)dx$$

    Soltn for Q2 for upper bounds

    $$\int_{0}^{1} \sqrt{1+2x+x^2}dx \\
    = \int_{0}^{1} \sqrt{(x+1)^2}dx \\
    U_{f}(P) = \int_{0}^{1} (x+1)dx$$


    Soltn for Q2 for lower bound: $$\int_{0}^{1}x^3h(x)dx \\

    = \int_{0}^{1}x^3(-4)dx \\
    L_{f}(P) = \int_{0}^{1}-4x^3dx$$

    Soltn for Q2 for upper bound....

    $$\int_{0}^{1}x^3(x^2-4)dx\\
    U_{f}(P) = \int_{0}^{1}x^6-4x^3dx$$


    Thank you
     
    Last edited by a moderator: Nov 10, 2015
  2. jcsd
  3. Nov 11, 2015 #2

    Samy_A

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    You have to check your computations.

    3 hints:
    For Q1, you seem to assume that ##-2x≤g(x)≤2x##, but that is not what the question states.
    When you compute a square root, you have to take the positive root.
    ##x^3x^2=x^5##
     
    Last edited: Nov 11, 2015
  4. Nov 11, 2015 #3

    Ray Vickson

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    In Q1, can't you just say that ##-2 \leq g(x) \leq 2##? How does it help to multiply everything by ##x \geq 0 ## to get ##-2x \leq x g(x) \leq 2x##?
     
  5. Nov 12, 2015 #4
    That was how the question was defined. It said: let g(x) be any continuous function that satisifes -2x≤g(x)≤2x for 0≤x≤1. Find the upper and lower bounds for ∫[from 0 to 1] √(1+g(x)+x^2)dx

    Thanks
     
  6. Nov 12, 2015 #5

    SammyS

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    Right.

    So the first line in the OP does have a typo.
    You had:
    ##\ -2x\le xg(x)\le 2x\ ##​

    It turns out it should be
    ##\ -2x\le g(x)\le 2x\ ##​

    This makes more sense.
     
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