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Math Help - Area under a curve

  1. #1
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    Area under a curve

    Help me, pls, to solve this task

    Find an area which limited y=((1-x)/(1+x))^(2/3), x=1, x=-1.
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  2. #2
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    Quote Originally Posted by Dext91 View Post
    Help me, pls, to solve this task

    Find an area which limited y=((1-x)/(1+x))^(2/3), x=1, x=-1.
    y = {\left( {\frac{{1 - x}}{{1 + x}}} \right)^{2/3}},{\text{ }}x =  - 1,{\text{ }}x = 1,{\text{ }}A=?

    A = \int\limits_{ - 1}^1 {{{\left( {\frac{{1 - x}}{{1 + x}}} \right)}^{2/3}}dx}  = \left\{ \begin{gathered}\frac{{1 - x}}<br />
{{1 + x}} = {t^3} \Leftrightarrow x = \frac{{1 - {t^3}}}<br />
{{1 + {t^3}}}, \hfill \\dx = \frac{{ - 6{t^2}}}<br />
{{{{\left( {1 + {t^3}} \right)}^2}}}dt \hfill \\ <br />
\end{gathered}  \right\} =

    =  - 6\int\limits_\infty ^0 {\frac{{{t^4}}}<br />
{{{{\left( {1 + {t^3}} \right)}^2}}}dt}  = 6\int\limits_0^\infty  {\frac{{{t^4}}}<br />
{{{{\left( {1 + {t^3}} \right)}^2}}}dt} .

    6\int {\frac{{{t^4}}}<br />
{{{{\left( {1 + {t^3}} \right)}^2}}}dt}  = 2\int {{t^2} \cdot \frac{{3{t^2}}}<br />
{{{{\left( {1 + {t^3}} \right)}^2}}}dt}  =  - 2\int {{t^2}d\left( {\frac{1}<br />
{{1 + {t^3}}}} \right)}  =

    =  - \frac{{2{t^2}}}<br />
{{1 + {t^3}}} + 4\int {\frac{t}<br />
{{1 + {t^3}}}dt} .


    \frac{t}{{1 + {t^3}}} = \frac{t}<br />
{{\left( {t + 1} \right)\left( {{t^2} - t + 1} \right)}} = \frac{A}<br />
{{t + 1}} + \frac{{Bt + C}}<br />
{{{t^2} - t + 1}} \Leftrightarrow

    \Leftrightarrow t = \left( {{t^2} - t + 1} \right)A + \left( {t + 1} \right)\left( {Bt + C} \right).

    t =  - 1:{\text{ }} - 1 = 3A \Leftrightarrow A =  - \frac{1}{3};

    t = 0:{\text{ }}0 = A + C =  - \frac{1}{3} + C \Leftrightarrow \frac{1}{3};

    t = 1:{\text{ }}1 = A + 2\left( {B + C} \right) =  - \frac{1}<br />
{3} + 2\left( {B + \frac{1}<br />
{3}} \right) \Leftrightarrow B = \frac{1}{3}.

    4\int {\frac{t}<br />
{{1 + {t^3}}}dt}  =  - \frac{4}<br />
{3}\int {\frac{{dt}}<br />
{{1 + t}}}  + \frac{4}<br />
{3}\int {\frac{{t + 1}}<br />
{{{t^2} - t + 1}}dt}  =

    =  - \frac{4}<br />
{3}\ln \left| {1 + t} \right| + \frac{2}<br />
{3}\int {\frac{{2t - 1 + 3}}<br />
{{{t^2} - t + 1}}dt}  =

    =  - \frac{4}<br />
{3}\ln \left| {1 + t} \right| + \frac{2}<br />
{3}\int {\frac{{2t - 1}}<br />
{{{t^2} - t + 1}}dt}  + 2\int {\frac{{dt}}<br />
{{{t^2} - t + 1}}}  =

    =  - \frac{4}<br />
{3}\ln \left| {1 + t} \right| + \frac{2}<br />
{3}\int {\frac{{d\left( {{t^2} - t + 1} \right)}}<br />
{{{t^2} - t + 1}}}  + 8\int {\frac{{dt}}<br />
{{4{t^2} - 4t + 1 + 3}}}  =

    =  - \frac{4}<br />
{3}\ln \left| {1 + t} \right| + \frac{2}<br />
{3}\ln \left| {{t^2} - t + 1} \right| + 8\int {\frac{{dt}}<br />
{{{{\left( {2t - 1} \right)}^2} + 3}}}  =

    =  - \frac{2}<br />
{3}\ln \left| {{t^2} + 2t + 1} \right| + \frac{2}<br />
{3}\ln \left| {{t^2} - t + 1} \right| + \frac{8}<br />
{3}\int {\frac{{dt}}<br />
{{{{\left( {\frac{{2t - 1}}<br />
{{\sqrt 3 }}} \right)}^2} + 1}}}  =

    = \frac{2}<br />
{3}\ln \left| {\frac{{{t^2} - t + 1}}<br />
{{{t^2} + 2t + 1}}} \right| + \frac{{4\sqrt 3 }}<br />
{3}\int {\frac{{d\left( {\frac{{2t - 1}}<br />
{{\sqrt 3 }}} \right)}}<br />
{{{{\left( {\frac{{2t - 1}}<br />
{{\sqrt 3 }}} \right)}^2} + 1}}}  =

    = \frac{2}<br />
{3}\ln \left| {\frac{{{t^2} - t + 1}}<br />
{{{t^2} + 2t + 1}}} \right| + \frac{{4\sqrt 3 }}<br />
{3}\arctan \frac{{2t - 1}}<br />
{{\sqrt 3 }} + C.

    So we have

    A = \left. {\left( { - \frac{{2{t^2}}}<br />
{{1 + {t^3}}} + \frac{2}<br />
{3}\ln \left| {\frac{{{t^2} - t + 1}}<br />
{{{t^2} + 2t + 1}}} \right| + \frac{{4\sqrt 3 }}<br />
{3}\arctan \frac{{2t - 1}}<br />
{{\sqrt 3 }}} \right)} \right|_0^\infty  =

    =  - 2\underbrace {\mathop {\lim }\limits_{t \to \infty } \frac{{{t^2}}}{{1 + {t^3}}}}_0 + \frac{2}{3}\underbrace {\mathop {\lim }\limits_{t \to \infty } \ln \left| {\frac{{{t^2} - t + 1}}<br />
{{{t^2} + 2t + 1}}} \right|}_0 + \frac{{4\sqrt 3 }}{3}\underbrace {\mathop {\lim }\limits_{t \to \infty } \arctan \frac{{2t - 1}}{{\sqrt 3 }}}_{\pi /2} + \frac{{4\sqrt 3 }}{3} \cdot \frac{\pi }{6} =

    = \frac{{4\sqrt 3 }}<br />
{3} \cdot \frac{\pi }<br />
{2} + \frac{{2\sqrt 3 }}<br />
{9}\pi  = \frac{{2\sqrt 3 }}<br />
{3}\pi  + \frac{{2\sqrt 3 }}<br />
{9}\pi  = \frac{{8\sqrt 3 }}<br />
{9}\pi .

    See this picture

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