# Thread: Area under a curve

1. ## 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.

2. Originally Posted by Dext91
Help me, pls, to solve this task

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

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

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

$\displaystyle 6\int {\frac{{{t^4}}} {{{{\left( {1 + {t^3}} \right)}^2}}}dt} = 2\int {{t^2} \cdot \frac{{3{t^2}}} {{{{\left( {1 + {t^3}} \right)}^2}}}dt} = - 2\int {{t^2}d\left( {\frac{1} {{1 + {t^3}}}} \right)} =$

$\displaystyle = - \frac{{2{t^2}}} {{1 + {t^3}}} + 4\int {\frac{t} {{1 + {t^3}}}dt} .$

$\displaystyle \frac{t}{{1 + {t^3}}} = \frac{t} {{\left( {t + 1} \right)\left( {{t^2} - t + 1} \right)}} = \frac{A} {{t + 1}} + \frac{{Bt + C}} {{{t^2} - t + 1}} \Leftrightarrow$

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

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

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

$\displaystyle t = 1:{\text{ }}1 = A + 2\left( {B + C} \right) = - \frac{1} {3} + 2\left( {B + \frac{1} {3}} \right) \Leftrightarrow B = \frac{1}{3}.$

$\displaystyle 4\int {\frac{t} {{1 + {t^3}}}dt} = - \frac{4} {3}\int {\frac{{dt}} {{1 + t}}} + \frac{4} {3}\int {\frac{{t + 1}} {{{t^2} - t + 1}}dt} =$

$\displaystyle = - \frac{4} {3}\ln \left| {1 + t} \right| + \frac{2} {3}\int {\frac{{2t - 1 + 3}} {{{t^2} - t + 1}}dt} =$

$\displaystyle = - \frac{4} {3}\ln \left| {1 + t} \right| + \frac{2} {3}\int {\frac{{2t - 1}} {{{t^2} - t + 1}}dt} + 2\int {\frac{{dt}} {{{t^2} - t + 1}}} =$

$\displaystyle = - \frac{4} {3}\ln \left| {1 + t} \right| + \frac{2} {3}\int {\frac{{d\left( {{t^2} - t + 1} \right)}} {{{t^2} - t + 1}}} + 8\int {\frac{{dt}} {{4{t^2} - 4t + 1 + 3}}} =$

$\displaystyle = - \frac{4} {3}\ln \left| {1 + t} \right| + \frac{2} {3}\ln \left| {{t^2} - t + 1} \right| + 8\int {\frac{{dt}} {{{{\left( {2t - 1} \right)}^2} + 3}}} =$

$\displaystyle = - \frac{2} {3}\ln \left| {{t^2} + 2t + 1} \right| + \frac{2} {3}\ln \left| {{t^2} - t + 1} \right| + \frac{8} {3}\int {\frac{{dt}} {{{{\left( {\frac{{2t - 1}} {{\sqrt 3 }}} \right)}^2} + 1}}} =$

$\displaystyle = \frac{2} {3}\ln \left| {\frac{{{t^2} - t + 1}} {{{t^2} + 2t + 1}}} \right| + \frac{{4\sqrt 3 }} {3}\int {\frac{{d\left( {\frac{{2t - 1}} {{\sqrt 3 }}} \right)}} {{{{\left( {\frac{{2t - 1}} {{\sqrt 3 }}} \right)}^2} + 1}}} =$

$\displaystyle = \frac{2} {3}\ln \left| {\frac{{{t^2} - t + 1}} {{{t^2} + 2t + 1}}} \right| + \frac{{4\sqrt 3 }} {3}\arctan \frac{{2t - 1}} {{\sqrt 3 }} + C.$

So we have

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

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

$\displaystyle = \frac{{4\sqrt 3 }} {3} \cdot \frac{\pi } {2} + \frac{{2\sqrt 3 }} {9}\pi = \frac{{2\sqrt 3 }} {3}\pi + \frac{{2\sqrt 3 }} {9}\pi = \frac{{8\sqrt 3 }} {9}\pi .$

See this picture