Math, asked by Anonymous, 2 days ago


  \sf{Evaluate \:  the \:  definite \:  integral, if \:  it \:  exists:}
 \displaystyle{  \mathrm{\int \limits_{ - 1}^{1} \dfrac{x + 1}{(x + 2)^{4} }  dx}}

Answers

Answered by MrCarnage
56

A symmetric function, since it has the form  \displaystyle \int \limits_ { - a}^{a} f( x)dx. However, the function  f(x) =   \mathrm{\dfrac{x + 1}{( {x}^{2} + 2)^{4} }} is neither odd, nor even, so it is not symmetric. We can try to use the substitution rule to solve this problem Let u = x + 2. Differentiating each side, we have.

du = dx

The substitution rule appears bas though it may not work, Since we do not end up with a term that cancels (x + 1) out of the integral. To solve this integral the Equation u = x + 2 as x = u - 2. Before we substitute the equation into the integral, we must change the limits to terms of u. When x = 1, u = 3 and when x = -1, u = 1.

\rule{200pt}{}

 \displaystyle{ \mathrm{\int \limits_{1}^{3} \dfrac{x + 1}{(x + 2)^{4} } dx} = \mathrm{\int \limits_{1}^{3} \dfrac{ \big(u - 2 \big) + 1}{ \big((u -  2)  + 2\big)^{4} }  \big(du \big)}}\\\\

  :  \longmapsto \displaystyle {\mathrm{ \int \limits _  {1}^{3} \frac{u - 1}{ {u}^{4} } du }}\\\\

  :  \longmapsto \displaystyle {\mathrm{ \int \limits _  {1}^{3} \frac{u }{  {u}^{4} }  -  \frac{1}{{u}^{4}  }du }}\\\\

  :  \longmapsto \displaystyle {\mathrm{ \int \limits _  {1}^{3} \frac{ \cancel{u }}{ \cancel{  {u}^{4}} }  -  \frac{1}{{u}^{4}  }du }}\\\\

  :  \longmapsto \displaystyle {\mathrm{ \int \limits _  {1}^{3} \frac{1}{  {u}^{3} }  -  \frac{1}{{u}^{4}  }du }}\\\\

  : \longmapsto \displaystyle{ \mathrm{ \int \limits _ {1}^{3}  {u}^{ - 3} -  {u}^{ - 4}  du}}\\\\

 :  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ \dfrac{ {u}^{ - 3 + 1} }{ - 3 + 1}  -  \dfrac{ {u}^{ - 4 + 1} }{ - 4 + 1}} \\  \\ \end{array}\right]_{1}^{3}\\\\

:  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ \dfrac{ {u}^{ - 2} }{ - 2}  -  \dfrac{ {u}^{ - 3} }{ -3 }} \\  \\ \end{array}\right]_{1}^{3}\\\\

:  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 1 }{  {2u}^{2} }   +  \dfrac{ {1} }{  {3u}^{3}  }} \\  \\ \end{array}\right]_{1}^{3}\\\\

:  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 1 }{  {2(3)}^{2} }   +  \dfrac{ {1} }{  {3(3)}^{3}  }} \\  \\ \end{array}\right] -\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 1 }{  {2(1)}^{2} }   +  \dfrac{ {1} }{  {3(1)}^{3}  }} \\  \\ \end{array}\right] \\\\

:  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 1 }{  18}   +  \dfrac{ {1} }{  81 }} \\  \\ \end{array}\right] -\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 1 }{  2 }   +  \dfrac{ {1} }{ 3 }} \\  \\ \end{array}\right] \\\\

:  \longmapsto\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 9}{ 162}   +  \dfrac{ 2 }{ 162 }} \\  \\ \end{array}\right] -\left[\begin{array}{c c c} \\  \mathrm{ -  \dfrac{ 3}{  6 }   +  \dfrac{ 2}{ 6 }} \\  \\ \end{array}\right] \\\\

 :  \longmapsto   -  \rm \dfrac{7}{162}   +  \dfrac{1}{6}\\\\

 :  \longmapsto   -  \rm \dfrac{7}{162}   +  \dfrac{27}{162}\\\\

 :  \longmapsto    \rm  \cancel\dfrac{20}{162}  \\\\

 :  \longmapsto    \rm  \dfrac{10}{81}  \\\\

The Definite integral  \displaystyle{ \mathrm{\int \limits_{ - 1}^{1} \dfrac{x + 1}{(x + 2)^{4} } dx}}\: is equal to \rm  \dfrac{10}{81}

 \rule{200pt}{7pt}

Answered by Anonymous
39

Refer the attachment for your answer :)

Note :-

I first calculated the respective indefinite integral , then apply the limits (:

If it doesn't help you , kindly report it :)

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