Amaths integration ):
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Hello, anyone who is online now can help me? ^^ You see, let's say I have a question:
Integrate from 3 to 4 for 1/(x-2)^3 dx.
It will be:
Integrate 3 to 4 for (x-2)^-3 dx
= [(x-2)^-2 / -2] 3 to 4
= solve etc.
But I don't understand, by adding 1 to the power and dividing the expression by the new power, we are only integrating the entire expression on itself right? Why don't we have to divide the bolded part by (x^2/2 - 2x) which is obtained from integrating (x-2) itself? In differentiation, I thought we have to differentiate the whole thing on itself and then the separate elements inside too? =[ This is the part I'm super confused on integration, because sometimes I don't divide it and I get it wrong and sometimes I divide it I also get it wrong.I hope you guys get what I'm trying to say. T.T I've tried to make it as clear as possible already. Thanks sooo much!
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differentiate your answer.
if you get the question back, it works.
integrate is just opposite of differentiate, don't think too deep.
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where ish �鹰？？？
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But then why is it when I integrate (2+3x)^5 the solutions are:
{ [(2+3x)^6]/6 } / 3
Divided by 3.. I assume it's from 3x? >.< I don't know.. =[ -
it is from 3x.
differentiate the answer, you will need to multiply by 3 because of 3x too right?
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Originally posted by skythewood:
it is from 3x.
differentiate the answer, you will need to multiply by 3 because of 3x too right?
So means, everytime if I divide I'll divide what I integrated with the coefficient of x and not the whole expression over again? Just like integrating 1/(x-2)^3 whereby I actually divide by 1, but because it'll be the same so there is no difference to the answer?
Originally posted by maurizio13:integrate 1/x result is always ln |x|
Yes, that's another question. When do I make it In =[ Alamak, I feel so dumb. -.-"
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Originally posted by iamapebble:
So means, everytime if I divide I'll divide what I integrated with the coefficient of x and not the whole expression over again? Just like integrating 1/(x-2)^3 whereby I actually divide by 1, but because it'll be the same so there is no difference to the answer?
Yes, that's another question. When do I make it In =[ Alamak, I feel so dumb. -.-"
yes, you get the idea.
There is no real integrating... there is only a opposite of differentiating..
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Hmmmm okayyy, thanks I'll go try more questions I guess.. =/
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i am very rusty, but i think it's because if the power is -1, there are no anti-derivatives.
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x^-1
integrate will get lnx
this is no technique in this integration, just memorize it.
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Originally posted by skythewood:
x^-1
integrate will get lnx
this is no technique in this integration, just memorize it.
There is, but it is out of O level syllabus.
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Originally posted by skythewood:
yes, you get the idea.
There is no real integrating... there is only a opposite of differentiating..
the concept of integration is different from differentiation... not fully opposite...
integration is sort of a type of summation, differentiation is a little of finding small differencesthen again, out of O level sylalbus :D
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O Level just memorise integrate x^-1 = ln x
Actually when do we need the mod?
ie integrate x^-1 = ln |x|
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Originally posted by secretliker:
O Level just memorise integrate x^-1 = ln x
Actually when do we need the mod?
ie integrate x^-1 = ln |x|
everytime need... better to put on the safe side
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My school never say MUST put leh...
I only learned "integrate 1/x = ln x". That's all. The mod is used to ensure that x > 0 right, since ln x where x < 0 is undefined.
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The mod is used to ensure that x > 0 right, since ln x where x < 0 is undefined.
Yup
I was taught in sec sch to put...
Anyway, put won't go wrong. Don't put might go wrong... So I think, put better :D
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I have no idea what's with the mod and no mod. -catch no ball- o.o
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because you cannot ln a negative number
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OHHHHHH I SEE I SEE! THANKS :D
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Originally posted by iamapebble:
Hello, anyone who is online now can help me? ^^ You see, let's say I have a question:
Integrate from 3 to 4 for 1/(x-2)^3 dx.
It will be:
Integrate 3 to 4 for (x-2)^-3 dx
= [(x-2)^-2 / -2] 3 to 4
= solve etc.
But I don't understand, by adding 1 to the power and dividing the expression by the new power, we are only integrating the entire expression on itself right? Why don't we have to divide the bolded part by (x^2/2 - 2x) which is obtained from integrating (x-2) itself? In differentiation, I thought we have to differentiate the whole thing on itself and then the separate elements inside too? =[ This is the part I'm super confused on integration, because sometimes I don't divide it and I get it wrong and sometimes I divide it I also get it wrong.I hope you guys get what I'm trying to say. T.T I've tried to make it as clear as possible already. Thanks sooo much!
So if your coefficient of a is 1, you would find that you only end up having (n+1) at the denominator.
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Originally posted by eagle:
the concept of integration is different from differentiation... not fully opposite...
integration is sort of a type of summation, differentiation is a little of finding small differencesthen again, out of O level sylalbus :D
hmmm... should i put "edited so as to not confuse students"?
:)
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Originally posted by skythewood:
hmmm... should i put "edited so as to not confuse students"?
:)
stated right there in case you missed it
out of O level sylalbus
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I saw, but same thoughts.
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Originally posted by skythewood:
I saw, but same thoughts.
well, to prove that it is confusing students, you have to prove it is wrong to do so within the A level syllabus.
don't be an idiot again (taking things like synergy to include into maths calculations, no such thing for pri sch, O levels or A levels), then insist you are correct.