I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8
But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure.
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Last edited by User947387; 14-01-2015 at 15:44. -
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(Original post by apronedsamurai)
I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8
But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure. -
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- 14-01-2015 17:22
(Original post by apronedsamurai)
I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8
But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure. -
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- 14-01-2015 20:19
r^3-2r^2-8r+c. Which integrates to;
1/4r^4-2/3r^3-4r^2+cLast edited by jshep000; 14-01-2015 at 20:20. -
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- 15-01-2015 14:38
Last edited by User947387; 15-01-2015 at 14:51.
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