Question

Qual a integral de : x^2*e^(-5x)?

Answer 1

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Calcular a integral indefinida:

$\large\begin{array}{l} \mathtt{\displaystyle\int\! x^2\cdot e^{-5x}\,dx}\\\\ \mathtt{\displaystyle\int\!\frac{1}{(-5)}\cdot (-5)\cdot x^2\cdot e^{-5x}\,dx}\\\\ \mathtt{\displaystyle\int\!\left(\!-\,\frac{1}{5}\right)x^2\cdot e^{-5x}\cdot (-5)\,dx\qquad\quad(i)} \end{array}$


$\large\begin{array}{l} \texttt{M\'etodo de integra\c{c}\~ao por partes:}\\\\ \begin{array}{lcl} \mathtt{u=-\,\dfrac{1}{5}\,x^2}&\quad\Rightarrow\quad&\mathtt{du=-\,\dfrac{2}{5}\,x\,dx}\\\\ \mathtt{dv=e^{-5x}\cdot (-5)\,dx}&\quad\Leftarrow\quad&\mathtt{v=e^{-5x}} \end{array}\end{array}$


$\large\begin{array}{l} \mathtt{\displaystyle\int\!u\,dv=u\cdot v-\int\! v\,du}\\\\ \mathtt{\displaystyle\int\!\left(\!-\,\frac{1}{5}\right)x^2\cdot e^{-5x}\cdot (-5)\,dx=-\,\frac{1}{5}\,x^2\cdot e^{-5x}-\int\! e^{-5x}\cdot \left(\!-\,\frac{2}{5}\right)x\,dx}\\\\ \mathtt{\displaystyle\int\!x^2\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x^2\cdot e^{-5x}+\frac{2}{5}\int\! x\cdot e^{-5x}\,dx}\\\\ \mathtt{\displaystyle\int\!x^2\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x^2\cdot e^{-5x}+\frac{2}{5}\cdot I_1\qquad\quad(ii)} \end{array}$


$\large\begin{array}{l} \textsf{onde}\\\\ \mathtt{I_1=\displaystyle\int\! x\cdot e^{-5x}\,dx}\\\\ \mathtt{I_1=\displaystyle\int\! \left(\!-\,\frac{1}{5}\right)x\cdot e^{-5x}\cdot (-5)\,dx}. \end{array}$

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$\large\begin{array}{l} \texttt{Para integrar }\mathtt{I_1}\textsf{, usamos o m\'etodo de integra\c{c}\~ao por partes}\\\texttt{novamente:}\\\\ \begin{array}{lcl} \mathtt{u=-\,\dfrac{1}{5}\,x}&\quad\Rightarrow\quad&\mathtt{du=-\,\dfrac{1}{5}\,dx}\\\\ \mathtt{dv=e^{-5x}\cdot (-5)\,dx}&\quad\Leftarrow\quad&\mathtt{v=e^{-5x}} \end{array} \end{array}$


$\large\begin{array}{l} \mathtt{\displaystyle\int\!u\,dv=u\cdot v-\int\! v\,du}\\\\ \mathtt{\displaystyle\int\!\left(\!-\,\frac{1}{5}\right)x\cdot e^{-5x}\cdot (-5)\,dx=-\,\frac{1}{5}\,x\cdot e^{-5x}-\int\! e^{-5x}\cdot \left(\!-\,\frac{1}{5}\right)\,dx}\\\\ \mathtt{\displaystyle\int\! x\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x\cdot e^{-5x}+\frac{1}{5}\int\!e^{-5x}\,dx}\\\\ \mathtt{\displaystyle\int\! x\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x\cdot e^{-5x}+\frac{1}{5}\int\!\left(\!-\,\frac{1}{5}\right)\cdot e^{-5x}\cdot (-5)\,dx} \end{array}$

$\large\begin{array}{l} \mathtt{\displaystyle I_1=-\,\frac{1}{5}\,x\cdot e^{-5x}-\frac{1}{25}\int\!e^{-5x}\cdot (-5)\,dx}\\\\ \mathtt{\displaystyle I_1=-\,\frac{1}{5}\,x\cdot e^{-5x}-\frac{1}{25}\int\!e^{w}\,dw\qquad(w=-5x)}\\\\ \mathtt{\displaystyle I_1=-\,\frac{1}{5}\,x\cdot e^{-5x}-\frac{1}{25}\,e^{w}}\\\\ \mathtt{\displaystyle I_1=-\,\frac{1}{5}\,x\cdot e^{-5x}-\frac{1}{25}\,e^{-5x}\qquad\quad(iii)} \end{array}$


$\large\begin{array}{l} \texttt{Substituindo de volta em (ii), a integral fica}\\\\ \mathtt{\displaystyle\int\!x^2\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x^2\cdot e^{-5x}+\frac{2}{5}\cdot \left(-\,\frac{1}{5}\,x\cdot e^{-5x}-\frac{1}{25}\,e^{-5x}\right)+C}\\\\ \mathtt{\displaystyle\int\!x^2\cdot e^{-5x}\,dx=-\,\frac{1}{5}\,x^2\cdot e^{-5x}-\frac{2}{25}\,x\cdot e^{-5x}-\frac{2}{125}\,e^{-5x}+C} \end{array}\\\\\\\\ \therefore~~\large\boxed{\begin{array}{c}\mathtt{\displaystyle\int\!x^2\cdot e^{-5x}\,dx=-\,\frac{1}{125}\cdot (25x^2+10x+2)\cdot e^{-5x}+C}\end{array}}\qquad\quad\checkmark$


Bons estudos! :-)


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