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:) 

 

 

 

http://prntscr.com/j35u6u

 

Someone else asked this question, but I'm also confused, could someone explain how this integral was obtained. I don't understand where the 1- part comes from

quilly  Apr 10, 2018
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http://prntscr.com/j35u6u


Someone else asked this question, but I'm also confused, could someone explain how this integral was obtained.
I don't understand where the 1- part comes from

\(\mathbf{\displaystyle 4\int \limits_{0}^{D} d\sigma\ \sigma^2\ e^{-2\sigma} = \ ?}\)

 

1. Apply Integration By Parts:
Formula:

 

\(\text{Let $u = \sigma^2$ } \qquad u' = 2\sigma \\ \text{Let $v' = e^{-2\sigma}$ } \qquad v = \int d\sigma \ e^{-2\sigma} = -\frac{1}{2}e^{-2\sigma}\)

\(\begin{array}{|rcll|} \hline \int \limits_{0}^{D} d\sigma\ \underbrace{\sigma^2}_{=u}\ \underbrace{e^{-2\sigma}}_{=v'} &=& \left[ \underbrace{\sigma^2}_{=u} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \right]_{0}^{D} -\int \limits_{0}^{D} d\sigma\ \underbrace{2\sigma}_{=u'} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \\\\ &=& \left[ -\frac{1}{2}e^{-2\sigma}\sigma^2\right]_{0}^{D} +\int \limits_{0}^{D} d\sigma\ e^{-2\sigma}\sigma \\\\ &=& -\frac{1}{2}e^{-2D}D^2 +\int \limits_{0}^{D} d\sigma\ e^{-2\sigma}\sigma \\ \hline \end{array}\)

 

2. Apply Integration By Parts:

\(\text{Let $u = \sigma$ } \qquad u' = 1 \\ \text{Let $v' = e^{-2\sigma}$ } \qquad v = \int d\sigma \ e^{-2\sigma} = -\frac{1}{2}e^{-2\sigma}\)

\(\begin{array}{|rcll|} \hline \int \limits_{0}^{D} d\sigma\ \underbrace{e^{-2\sigma}}_{=v'} \underbrace{\sigma}_{=u} &=& \left[ \underbrace{\sigma}_{=u} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \right]_{0}^{D} -\int \limits_{0}^{D} d\sigma\ \underbrace{1}_{=u'} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \\\\ &=& D \left( -\frac{1}{2}e^{-2D}\right) +\int \limits_{0}^{D} d\sigma\ \frac{1}{2}e^{-2\sigma} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \int \limits_{0}^{D} d\sigma\ e^{-2\sigma} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2\sigma} \right]_{0}^{D} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}e^{-2\cdot 0})\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}e^{0})\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}\cdot 1)\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D}+\frac{1}{2} \right] \\\\ &=& -\frac{1}{2} e^{-2D}D - \frac{1}{4} e^{-2D} + \frac{1}{4} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{4\int \limits_{0}^{D} d\sigma\ \sigma^2 e^{-2\sigma} } &=& 4\left[ -\frac{1}{2}e^{-2D}D^2 -\frac{1}{2} e^{-2D}D - \frac{1}{4} e^{-2D} + \frac{1}{4} \right] \\\\ &=& -2e^{-2D}D^2 -2e^{-2D}D - e^{-2D} +1 \\\\ &=& -e^{-2D}(2D^2 +2D +1) +1 \\\\ &\mathbf{=}& \mathbf{1-e^{-2D}(2D^2 +2D +1) } \\ \hline \end{array}\)

 

 

laugh

heureka  Apr 10, 2018
 #2
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i agree with heureka

jakesplace  Apr 10, 2018

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