anonymous
  • anonymous
Integrate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
This is an integration by parts problem, but fortunately there is a trick to reduce it a bit:
anonymous
  • anonymous
\[\int\limits x^2 e^{-x/4} dx = \int\limits (4u)^2 e^{-u} d(4u) = 4^3 \int\limits u^2 e^{-u} du \]

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anonymous
  • anonymous
Where I used the substitution u = x / 4 so that x = 4u and dx = d(4u) = 4 du
anonymous
  • anonymous
Now let me just call everything y instead so I can use the u in the typical form of integration by parts: \[\int\limits d(uv) = \int\limits u dv + \int\limits vdu\] Which if you drop the integral signs is merely the Liebniz rule for calculating derivatives: \[\frac{ d }{ dx } (uv) = u \frac{ d }{ dx }v + v \frac{ d }{ dx }u\] Manipulating this integral term gives: \[ \int\limits u dv = uv - \int\limits v du\] So lets take: \[4^3 \int\limits y^2 e^{-y} dy\] where again I changed the name of the original subsitution variable to y=x/4 and get to work
anonymous
  • anonymous
I used integ by parts and I got
anonymous
  • anonymous
|dw:1442185882375:dw|
anonymous
  • anonymous
Wow that was wierd the page froze....
anonymous
  • anonymous
\[\int\limits x^2 e^{-x/4} dx = 4^3 \int\limits y^2 e^{-y} dy\] where we choose u=y^2 and dv = e^-y So: \[ \int\limits y^2 e^{-y} dy = -y^2 e^{-y} + 2 \int\limits y e^{-y} dy \]
anonymous
  • anonymous
Thank you so much =)
anonymous
  • anonymous
Now we choose u = y and dv = e^-y (again) which gives: \[ \int\limits y e^{-y} dy = -ye^{-y} + \int e^{-y} dy = -ye^{-y} - e^{-y} \] Putting It all together: \[ \int x^2 e^{-x/4} dx = 4^3\int\limits y^2 e^{-y} dy =4^3 ( -y^2 e^{-y} + 2 \int\limits y e^{-y} dy ) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4^3 ( -y^2 e^{-y} + 2 (-ye^{-y} - e^{-y})) = -64e^{-y}(y^2 -2 y + 1) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -64e^{-y}(y-1)^2\]
anonymous
  • anonymous
Now dont forget to substitute back in y=x/4 so your answer is in terms of the original function
anonymous
  • anonymous
\[ \int x^2 e^-{x/4} dx = -64e^{-x/4}( \frac{x}{4}- 1)^2 \]
anonymous
  • anonymous
Well I messed up the exponent but you get the point
anonymous
  • anonymous
Hey how did you get that human calculator tag btw?
anonymous
  • anonymous
The smartscore thing..
anonymous
  • anonymous
Ahhh yea I am still relatively new here, but, as you can see, I do believe I am eligible for such a title :D :D
anonymous
  • anonymous
I just need more medals
anonymous
  • anonymous
Anyways see you around

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