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Math Help - Definite integral

  1. #1
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    Definite integral

    How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!
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  2. #2
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    Re: Definite integral

    Quote Originally Posted by Hamzawesome View Post
    How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!
    Put u=\sqrt{x} then simplify and use partial fractions.

    CB
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  3. #3
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    Re: Definite integral

    Quote Originally Posted by Hamzawesome View Post
    How do I find the definite integral of 1 / (x^1/2 + 2x + x^3/2) dx, with upper limit of 4 and lower limit of 0? The hint says that I need to factor out the denominator, with which I got x^1/2 times (1 + 2x^1/2 + x). What do I do from here, and would u/du substitution be the best way? Thanks!
    \sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt

    I=2\cdot\int\limits_{0}^{2} \frac{1}{t^2+2t+1} \, dt =2\cdot\int\limits_{0}^{2} \frac{1}{(t+1)^2} \, dt
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    Re: Definite integral

    Quote Originally Posted by princeps View Post
    \sqrt{x}=t \Rightarrow \frac{1}{2\sqrt{x}}dx=dt

    I=2\cdot\int\limits_{0}^{2} \frac{1}{t^2+2t+1} \, dt =2\cdot\int\limits_{0}^{2} \frac{1}{(t+1)^2} \, dt
    Why did you put for the interval of the integral (2,0)? In the problem the upper limit is 4 and the lower limit is 0, did you put 2 and 0 for the limits by mistake or was that supposed to be there?
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    Re: Definite integral

    Quote Originally Posted by Hamzawesome View Post
    Why did you put for the interval of the integral (2,0)? In the problem the upper limit is 4 and the lower limit is 0, did you put 2 and 0 for the limits by mistake or was that supposed to be there?
    Since substitution is \sqrt {x}=t you have to change limits .

    \sqrt{0}=0 and \sqrt{4}=2
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    Re: Definite integral

    Quote Originally Posted by princeps View Post
    Since substitution is \sqrt {x}=t you have to change limits .

    \sqrt{0}=0 and \sqrt{4}=2
    After that I used u/du substitution for that last integral you included; u = t + 1 and du = dt, so the integral was 2int(2,0) du/u^2 which ended up as [-2/u](2,0) and then [-2/t+1](2,0) so for the final answers I got 4/3.
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    Re: Definite integral

    Quote Originally Posted by Hamzawesome View Post
    After that I used u/du substitution for that last integral you included; u = t + 1 and du = dt, so the integral was 2int(2,0) du/u^2 which ended up as [-2/u](2,0) and then [-2/t+1](2,0) so for the final answers I got 4/3.
    That should be a correct answer...
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  8. #8
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    Re: Definite integral

    Quote Originally Posted by princeps View Post
    That should be a correct answer...
    I hope you don't mean there are more than one....lol. Thanks, regardless.
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  9. #9
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    Re: Definite integral

    interesting to note that the original definite integral is improper ...
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