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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    Does anyone know how to do these problems. I'm stuck on them, and I don't know where to start. Thanks for any help.

    1. the integral from 0 to 1 of (theta + 1)/ the square root of ( theta^2 + 2 theta)

    2. the integral from 0 to infinty of (16tangent inverse of x)/(1 + x^2) dx

    3. the integral from negative infinty to infinity of 2xe^(-xsquared) dx
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  2. #2
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    Quote Originally Posted by turtle View Post
    1. the integral from 0 to 1 of (theta + 1)/ the square root of ( theta^2 + 2 theta)
    \int_0^1 \frac{x+1}{\sqrt{x^2+2x}} dx
    This is a type two improper integral for it is not defined at a point in the interval.

    We will find,
    \lim_{t\to 0^+} \int_t^1 \frac{x+1}{\sqrt{x^2+2x}} dx
    You can find,
    \int \frac{x+1}{\sqrt{x^2+2x}} dx
    By using the inner function,
    u=x^2+2x
    Thus,
     \frac{1}{2} \int \frac{1}{\sqrt{u}} du=\frac{1}{2} \cdot \frac{2}{1} u^{1/2} +C=\sqrt{x^2+2x}+C
    Thus,
    \lim_{t\to 0^+}\sqrt{1^2+2(1)}-\sqrt{t^2+2t}=\sqrt{3}
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  3. #3
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    Quote Originally Posted by turtle View Post

    2. the integral from 0 to infinty of (16tangent inverse of x)/(1 + x^2) dx
    \int_0^{\infty} \frac{16 \tan^{-1} x}{1+x^2} dx
    To find,
    \int \frac{16 \tan^{-1} x}{1+x^2} dx
    Use the inner function u=\tan^{-1} x.
    To get,
    \int 16 u du=8u^2+C=8(\tan^{-1} x)^2 +C
    Thus, the integral is,
    \lim_{t\to \infty}8 (\tan^{-1} t)^2 - 8(\tan^{-1} 0)^2=8\cdot \frac{\pi^2}{4}=2\pi^2
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  4. #4
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    Quote Originally Posted by turtle View Post
    3. the integral from negative infinty to infinity of 2xe^(-xsquared) dx
    Subdivide,
    \int_{-\infty}^0 2xe^{-x^2} dx+\int_0^{\infty} 2xe^{-x^2}dx
    The integral,
    \int 2xe^{-x^2} dx=e^{-x^2}+C
    Thus,
    \lim_{t\to -\infty}e^{-0^2}-e^{-t^2}+\lim_{t\to \infty} e^{-t^2}-e^{-0^2} =0
    Of you can use the fact that,
    y=2xe^{-x^2} is odd.
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  5. #5
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    Hello, turtle!

    These require only simple substitutions . . .


    1)\;\;\int^1_0\frac{\theta + 1}{\sqrt{\theta^2 + 2\theta}}\,d\theta

    We have: . \int(\theta^2 + 2\theta)^{-\frac{1}{2}}(\theta + 1)\,d\theta

    Let: u = \theta^2 + 2\theta\quad\Rightarrow\quad du = (2\theta + 2)d\theta\quad\Rightarrow\quad d\theta = \frac{du}{2(\theta + 1)}

    Substitute: . \int u^{-\frac{1}{2}}(\theta + 1)\frac{du}{2(\theta + 1)} \;=\;\frac{1}{2}\int u^{-\frac{1}{2}}\,du \;=\;u^{\frac{1}{2}} \;=\;\sqrt{u}

    Back-substitute: . \sqrt{\theta^2 + 2\theta}\,\bigg|^1_0\;=\;\sqrt{1 + 2} - \sqrt{0 + 0} \:=\:\sqrt{3}



    2)\;\;\int_0^{\infty} \frac{16\tan^{-1}x}{1 + x^2}\,dx

    We have: . 16\int^{\infty}_0\tan^{-1}x\cdot\frac{dx}{1 + x^2}

    Let  u = \tan^{-1}x\quad\Rightarrow\quad du = \frac{dx}{1 + x^2}

    Substitute: . 16\int^{\infty}_0u\:du \;=\;8u^2

    Back-substitute: . 8\left[\tan^{-1}x\right]^2\,\bigg|^{\infty}_0

    Evaluate: . 8\bigg[\left(\tan^{-1}(\infty)\right)^2 - \left(\tan^{-1}(0)\right)^2\bigg]\;=\;8\bigg[\left(\frac{\pi}{2}\right)^2 - 0^2\bigg]\;=\;2\pi^2

    Sheesh! . . . Too slow again . . .
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