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RickyOswaldIOW
2006-01-14 00:46:41

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.

Ricky
2006-01-14 00:39:15

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.

RickyOswaldIOW
2006-01-14 00:30:19

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?

RickyOswaldIOW
2006-01-13 22:46:48

sorry that my formatting is pretty c r a p p y
I think it should be
∫(2x^2 - x/2)dx = ...

Ricky
2006-01-13 22:44:17

∫2x^2 - x/2

∫2x^2 - ∫x/2

2∫x^2 - 1/2 ∫ x

2/3 x^3 - 1/4x^2 + C

RickyOswaldIOW
2006-01-13 21:55:07

∫dy/dx = ∫2x^2 - x/2
y = (2x^3)/3 - (x^2)/2(2) + C   =    (2x^3)/3 - (x^2)/4 + C

RickyOswaldIOW
2006-01-13 21:28:49

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

RickyOswaldIOW
2006-01-13 21:20:13

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