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The question is:
integrate the function 2x^2 - x/2
The first part would bt (2x^3)/3 but I am unsure of how to integrate the fraction.
*notices the bottom of the screen*
cool update
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I'll make a somewhat wild guess
∫dy/dx = ∫2x^2 - x/2
y = (2x^3)/3 - ((x^2)/2) / 2x + C
*edit - forgot the + c
Last edited by rickyoswaldiow (2006-01-12 22:34:08)
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∫dy/dx = ∫2x^2 - x/2
y = (2x^3)/3 - (x^2)/2(2) + C = (2x^3)/3 - (x^2)/4 + C
Last edited by rickyoswaldiow (2006-01-12 22:59:33)
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∫2x^2 - x/2
∫2x^2 - ∫x/2
2∫x^2 - 1/2 ∫ x
2/3 x^3 - 1/4x^2 + C
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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same answer
sorry that my formatting is pretty c r a p p y
I think it should be
∫(2x^2 - x/2)dx = ...
Last edited by rickyoswaldiow (2006-01-12 23:47:26)
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3/(4x^5) = 12x^-5
∫(12x^-5)dx = (12x^-4)/-4 + c = -12/4x^4 + c = -3x^4 + c
Is this a case of bad algebra or am I taking a wrong step in my integration?
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Bad algebra. It's not (4x)^5, it's 4 (x)^5.
A good way to rewrite it is:
4^1 * x^5
So 3 / (4^1 * x^5) = 3*4^-1 * x^-5. Of course, this is just (3/4) x^-5.
Edit: Oh, and my point in my first post was that you should just take the fraction out before you integrate, then multiply. I forgot to say that though.
Last edited by Ricky (2006-01-13 01:40:58)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Since I did my GCSE mathematics at a state school about 3 years ago (I skieved off or was stoned all the time), I remember nothing of it. Starting at Advanced GCSE level now means there's a lot that I'm expected to know that I don't
Do you know any good resources for learning algebra? I'm sure if I revised for a week I'd stop making (as many) silly mistakes.
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