Seed5813
New member
- Joined
- Jan 29, 2018
- Messages
- 24
1. LarCalcET6 5.8.037.
Find or evaluate the integral by completing the square:
. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{\sqrt{\strut -x^2\, -\, 12x\,}}\, dx\)
So when I did complete the square I did
-x^2 -12x
= -(x^2 + 12x)
= -(x^2 +12x +36) - 36
= -36 - (x + 6)^2
I didn't know what to do with the negative so I just ignored it and got the right answer but I feel like I cheated. That negative in front of the x is what F'd me up.
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Find or evaluate the integral by completing the square:
. . . . .\(\displaystyle \displaystyle \int\, \dfrac{1}{\sqrt{\strut -x^2\, -\, 12x\,}}\, dx\)
So when I did complete the square I did
-x^2 -12x
= -(x^2 + 12x)
= -(x^2 +12x +36) - 36
= -36 - (x + 6)^2
I didn't know what to do with the negative so I just ignored it and got the right answer but I feel like I cheated. That negative in front of the x is what F'd me up.
Sent from my LGLS755 using Tapatalk
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