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Math Help - Convexity and integrals

  1. #1
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    Convexity and integrals

    I need help in three basic problems. (But unfortunatelly I can't solve them!)

    1. Where is the following function convex/concave? +infection points
    y=\frac{1}{\ln(x)}

    2.What is the integral of the following function?
    f(x)=\frac{\sqrt{1+x^2}}{x}

    3.Determine the surface of revolutionif we rotate the function around the x-axis!
    y=e^{-x}, \left[0,\infty\right[

    Thank you very much in advance!
    Happy new year!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by doug View Post
    1. Where is the following function convex/concave? +infection points
    y=\frac{1}{\ln(x)}
    Study the sign of y''.

    Fernando Revilla
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by doug View Post
    3.Determine the surface of revolutionif we rotate the function around the x-axis!
    y=e^{-x}, \left[0,\infty\right[
    Find V=\pi\int_0^{+\infty}e^{-2x}\;dx

    Fernando Revilla
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by doug View Post
    2.What is the integral of the following function?
    f(x)=\frac{\sqrt{1+x^2}}{x}
    Use the substitution x=\tan t .

    Fernando Revilla
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  5. #5
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    Quote Originally Posted by doug View Post
    I need help in three basic problems. (But unfortunately I can't solve them!) ...

    2.What is the integral of the following function?
    f(x)=\frac{\sqrt{1+x^2}}{x}
    ...
    Thank you very much in advance!
    Happy new year!

    I found that for me, the substitution \displaystyle u=\sqrt{1+x^2} works well.

    In my opinion, the trig substitution, \displaystyle x=\tan(t)\,, gives a result which is still rather difficult to work with.
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  6. #6
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    Quote Originally Posted by doug View Post
    I need help in three basic problems. (But unfortunately I can't solve them!) ...

    3.Determine the surface of revolution if we rotate the function around the x-axis!
    y=e^{-x}, \left[0,\infty\right[

    Thank you very much in advance!
    Happy new year!

    I see that FernandoRevilla gave the formula for volume, not surface area.

    Surface Area, S, for a solid of revolution produced by revolving a suitable function, f(x), about the x-axis is: \displaystyle S=2\pi\int_a^b{f(x)\sqrt{1+\left(f'(x)\right)^2}}\  ,dx\,.

    For, \displaystyle f(x)=e^{-x}\,,\ \ S=2\pi\int_0^{\infty}{e^{-x}\sqrt{1+e^{-2x}}}\,dx\,.

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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by SammyS View Post
    I see that FernandoRevilla gave the formula for volume, not surface area.
    Right (I misread the question). Thanks.

    Fernando Revilla
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