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Math Help - [SOLVED] Verify an inequality/integral value

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Verify an inequality/integral value

    Without doing any calculus of an integral containing a sine, verify that \int_0^{\frac{\pi}{2}}xsin(x)\leq\frac{\pi^2}{8}. My idea was by bounding the integral to get something greater but still lesser than \frac{\pi^2}{8}. So I thought about bounding \int_0^{\frac{\pi}{2}}xsin(x) this way : \int_0^{\frac{\pi}{2}}xsin(x)\leq \int_0^{\frac{\pi}{2}}x=\frac{\pi}{2}, which is unfortunately greater than \frac{\pi^2}{8}. I have no other idea of how to procede. Any idea is welcome.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by arbolis View Post
    Without doing any calculus of an integral containing a sine, verify that \int_0^{\frac{\pi}{2}}xsin(x)\leq\frac{\pi^2}{8}. My idea was by bounding the integral to get something greater but still lesser than \frac{\pi^2}{8}. So I thought about bounding \int_0^{\frac{\pi}{2}}xsin(x) this way : \int_0^{\frac{\pi}{2}}xsin(x)\leq \int_0^{\frac{\pi}{2}}x=\frac{\pi}{2}, which is unfortunately greater than \frac{\pi^2}{8}. I have no other idea of how to procede. Any idea is welcome.
    Take a look at the graph. The red curve is x~sin(x) and the blue triangle is square. (The triangle is the area enclosed by the line y = x, the line y = 0 and x = pi / 2.)

    What is the area of the triangle?

    -Dan
    Attached Thumbnails Attached Thumbnails [SOLVED] Verify an inequality/integral value-sine.jpg  
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  3. #3
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    Quote Originally Posted by topsquark View Post
    What is the area of the triangle?
    Oh, I know, teacher pick me, no pick me, I know, can I say?

    It is, \frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{8}.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Take a look at the graph. The red curve is and the blue triangle is square. (The triangle is the area enclosed by the line y = x, the line y = 0 and x = pi / 2.)

    What is the area of the triangle?
    If I'm not wrong, it's \frac{\pi^2}{8}! Nice graphic. It verifies graphically the relation we must verify. Now I'm asking if it's possible to verify it algebraically or any other way that respect the condition "Without doing any calculus of an integral containing a sine", or course not a cosine, or something similar. (I'm asking this because I'm more than sure it's what the expect from us to do).
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by arbolis View Post
    If I'm not wrong, it's \frac{\pi^2}{8}! Nice graphic. It verifies graphically the relation we must verify. Now I'm asking if it's possible to verify it algebraically or any other way that respect the condition "Without doing any calculus of an integral containing a sine", or course not a cosine, or something similar. (I'm asking this because I'm more than sure it's what the expect from us to do).
    The proof is in post #1. The only mistake is that an anti-derivative of x is \frac{x^2}{2} rather than x.
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  6. #6
    MHF Contributor arbolis's Avatar
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    The proof is in post #1. The only mistake is that an anti-derivative of is rather than .
    Ahahahaha!!!!!!!!! You're right, what an error from my part!!! Thanks.
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