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Math Help - Integral inequality

  1. #1
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    Integral inequality

    Prove that:
    1/(3 square root(2)) less than or equal to integral from a to b of x^2/(square root (1 + x^2)) dx less than or equal to 1/3
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Coda202 View Post
    Prove that:
    1/(3 square root(2)) less than or equal to integral from a to b of x^2/(square root (1 + x^2)) dx less than or equal to 1/3
    There are not restrictions on a and b?

    and is this your inequality?

    \frac{1}{3\sqrt{2}}\leqslant\int_a^{b}\frac{x^2}{\  sqrt{1+x^2}}\leqslant\frac{1}{3}
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  3. #3
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    Integral Inequality

    Ahh I do apologize; [a,b] = [0,1], and yes that is the inequality which I meant.
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  4. #4
    Moo
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    Yop,

    \forall x \in [0,1],~ 1+x^2 \geq 1 \implies \frac{1}{\sqrt{1+x^2}} \leq 1

    And \forall x \in [0,1],~ 1+x^2 \leq 1+1^2=2 \implies \frac 12 \leq \frac{1}{\sqrt{1+x^2}}

    Hence \frac{x^2}{2} \leq \frac{x^2}{\sqrt{1+x^2}} \leq x^2


    Therefore \int_0^1 \frac{x^2}{2}~ dx \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \int_0^1 x^2~ dx

    \frac 16 \leq \int_0^1 \frac{x^2}{\sqrt{1+x^2}}~ dx \leq \frac 13
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