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Math Help - Integral

  1. #1
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    Integral

    Hello, help me please with this integral :

    prove that : \left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq  x^{4}+ (2x-1)^{2}
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    Quote Originally Posted by linda2005 View Post
    Hello, help me please with this integral :

    prove that : \left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq x^{4}+ (2x-1)^{2}
    Note that t \sqrt{3 + \cos t} \leq t \sqrt{3 + 1} = 2t.
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  3. #3
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    Quote Originally Posted by linda2005 View Post
    Hello, help me please with this integral :

    prove that : \left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq x^{4}+ (2x-1)^{2}
    Well

    -1 \le \cos t \le 1 so 2 \le 3 + \cos t \le 4 so \sqrt{2} \le \sqrt{3 + \cos t }\le 2
    \int_{2x-1}^{x^2}\sqrt{2}t \,dt \le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le \int_{2x-1}^{x^{2}}2t\,dt

    \left.\frac{\sqrt{2}}{2}t^2\right|_{2x-1}^{x^2} \le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le \left. t^2\right|_{2x-1}^{x^2}

    \frac{\sqrt{2}}{2}\left( x^4 - (2x-1)^2\right)\le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le x^4 - (2x-1)^2\le x^4 + (2x-1)^2
    from which your inequality follows.
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