anonymous one year ago Integrate

1. anonymous

2. anonymous

This is an integration by parts problem, but fortunately there is a trick to reduce it a bit:

3. anonymous

$\int\limits x^2 e^{-x/4} dx = \int\limits (4u)^2 e^{-u} d(4u) = 4^3 \int\limits u^2 e^{-u} du$

4. anonymous

Where I used the substitution u = x / 4 so that x = 4u and dx = d(4u) = 4 du

5. 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

6. anonymous

I used integ by parts and I got

7. anonymous

|dw:1442185882375:dw|

8. anonymous

Wow that was wierd the page froze....

9. 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: $\int\limits y^2 e^{-y} dy = -y^2 e^{-y} + 2 \int\limits y e^{-y} dy$

10. anonymous

Thank you so much =)

11. 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: $\int x^2 e^{-x/4} dx = 4^3\int\limits y^2 e^{-y} dy =4^3 ( -y^2 e^{-y} + 2 \int\limits y e^{-y} dy ) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4^3 ( -y^2 e^{-y} + 2 (-ye^{-y} - e^{-y})) = -64e^{-y}(y^2 -2 y + 1) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -64e^{-y}(y-1)^2$

12. anonymous

Now dont forget to substitute back in y=x/4 so your answer is in terms of the original function

13. anonymous

$\int x^2 e^-{x/4} dx = -64e^{-x/4}( \frac{x}{4}- 1)^2$

14. anonymous

Well I messed up the exponent but you get the point

15. anonymous

Hey how did you get that human calculator tag btw?

16. anonymous

The smartscore thing..

17. 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

18. anonymous

I just need more medals

19. anonymous

Anyways see you around