Integrals
- Thread starter Dazed
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- Apr 12, 2005
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The first one must just be scaring you.
Point #1, if you REALLY mean dz, then it's REALLY easy. Just add a 'z' and your constant of integration.
Point #2, if you meant dx, it's a little trickier, but you will like the simplicity of the solution. Just guess! Think Polynomial Rule and Chain Rule. If it were [int] 1/(x^4) dx, what would you do?
Point #1, if you REALLY mean dz, then it's REALLY easy. Just add a 'z' and your constant of integration.
Point #2, if you meant dx, it's a little trickier, but you will like the simplicity of the solution. Just guess! Think Polynomial Rule and Chain Rule. If it were [int] 1/(x^4) dx, what would you do?
- Joined
- Apr 12, 2005
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Well, that's discouraging.Dazed said:
It's just the polynomial rule.
[int] 1/(x^4) dx = [int] x^(-4) dx = x^(-3)/(-3) + C
[int] 1/[(2x+3)^4] dx = [int] (2x+3)^(-4) dx = (2x+3)^(-3)/[(-3)*2] + C
The Chain rule insists on the additional factor of '2'.
Do it a piece at a time.
[int] 4/t^5 dt + [int] 3[square root]t dt - [int] 4cos^2tdt