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#1 |
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FFR Player
Join Date: Mar 2009
Posts: 26
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I have just a couple questions regarding integrals
first question: if I'm taking the integral of something like: ∫x^5dx would it equal x^6/6+c without the integral sign? or is the integral sign included second question: I need help finding the integral of functions that include fractional exponents like: ∫8^t 2/3 dt = 8∫t^2/3dt = (8^t5/3)/5/3 +c (because I added one to the exponent) but I get stuck here.. what do I do after this? multiply by it's recipriol? I'm not sure how to do that Last question: is it true that: ∫1/xdx = natural log of the absolute value of x plus your constant? I want to make sure I did this problem correctly: ∫9/u du = 9∫1/u du = 9ln lul + c? or is this wrong? thx |
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#2 |
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slimy, yet ... satisfying
Join Date: May 2007
Age: 28
Posts: 1,244
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Preeetty sure everything you did there was correct, haha.
1 - no integral sign, that's all. 2 - is it 8^t2/3 or 8(t^2/3)? If it's the second, then the solved integral is just 8(3/5)t^5/3, or 24t^5/3 all over 5. 3 - correct. Hope that helps, even though you don't look like ya needed it. :P Last edited by Tidus810; 01-19-2011 at 07:43 PM.. |
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#3 |
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x'); DROP TABLE FFR;--
Join Date: Nov 2010
Posts: 6,334
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1. Without
2. Assuming you ask for the integral of 8(t^2/3), it'd be like 8 * the integral of t^2/3. Increasing the exponent here by 1 is the same as increasing it by 3/3, so (t^(5/3))/(5/3). Multiply by the 8 you moved out and you get (24/5)t^(5/3) + C 3. The integral of 1/x dx is ln(x)+C. So the integral of 9/u du is simply 9*ln(u) + C EDIT: Watch out Tidus -- you divide by the raised exponent, not multiply Last edited by Reincarnate; 01-19-2011 at 07:48 PM.. |
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#4 |
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FFR Player
Join Date: Mar 2009
Posts: 26
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perfect
thanks guys |
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