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how to integrate this? (1 Viewer)

nichhhole

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f3nr15 said:
1/ [20t-5t^2]

= [20t-5t^2] ^[-1]

by using ur rule wouldnt we add 1 to n which is [-1]
which would =0....

futhermore..
ax+b is a linear eq...[ie when you differentiate you get a]
wheras the eq we would be using isnt.. [20t-5t^2]...[if we diff here we would get 20-10t.. ]


bah. wdf do we do?
 

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chousta said:
wow.....so much effort on such a question........



hot tip: complete the square, and use a result such as invers tan...

well thats wat i would do...
I don't recall inverse trigonometry being in 2-Unit but that sounds like a good idea ...
 

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jannny said:
huh? wtf is inverse tan .

Can some1 post a completed workout?
If your teacher gave you that question, and you havn't even learnt trignometric functions yet i don't think thats how to do it. I havn't learnt that chapter either yet.
 

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nichhhole said:
1/ [20t-5t^2]

= [20t-5t^2] ^[-1]

by using ur rule wouldnt we add 1 to n which is [-1]
which would =0....

futhermore..
ax+b is a linear eq...[ie when you differentiate you get a]
wheras the eq we would be using isnt.. [20t-5t^2]...[if we diff here we would get 20-10t.. ]


bah. wdf do we do?
So far I think the integral should be a natural logarithm or an inverse trigonometric function ...
 

williamc

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nichhhole said:
bah. wdf do we do?
nfi



To the OP where did you get the question from? From what i have learnt so far in 2 unit, i'm pretty sure that isn't a 2 unit question, or the answer is that, that function can not be integrated.
 

Loctorak

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f3nr15 said:
I thought you integrated to a natural logarithm (loge f(x))
Isn't that only when the bottom has to do with the differentiation of the first? Or vice versa? In either case I don't think either have very much to do with the differentiation of each other. :)

I'm with the first answer, change it to a negative power and use the function of a function rule. Going from there should be easy.
 

collide

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I did this.. but I'm not sure :S it could be wrong

∫1/20t-5t^2 dt
= 1/20-10t x ∫(20-10t)/20t-5t^2 dt

=1/20-10t x ln (20t-5t^2)

=ln(20t-5t^2)/(20-10t ) + c
 

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Loctorak said:
Isn't that only when the bottom has to do with the differentiation of the first? Or vice versa? In either case I don't think either have very much to do with the differentiation of each other. :)

I'm with the first answer, change it to a negative power and use the function of a function rule. Going from there should be easy.
But - 1 + 1 = 0, you can't get a power it must become a natural logarithm instead.

You can use ∫ xn dx = xn+1/n+1 + C, IF n is not equal to 1

collide said:
I did this.. but I'm not sure :S it could be wrong

∫1/20t-5t^2 dt
= 1/20-10t x ∫(20-10t)/20t-5t^2 dt

=1/20-10t x ln (20t-5t^2)

=ln(20t-5t^2)/(20-10t ) + c
I was thinking the VERY same.
Then you could use the quotient rule and check if you are correct.
 

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collide said:
I did this.. but I'm not sure :S it could be wrong

∫1/20t-5t^2 dt
= 1/20-10t x ∫(20-10t)/20t-5t^2 dt

=1/20-10t x ln (20t-5t^2)

=ln(20t-5t^2)/(20-10t ) + c
I somewat recall my teacher saying you can't take out t or x or y.. because how can u integrate if it is outside the Integral

Oh btw, I got this q from my the school holidays homework.. and I forgot to get the answers sheet.
 

jb_nc

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this is the answer

Code:
-Log[-4 + x] + Log[x]
---------------------
         20
 

williamc

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The notation in the original question showed the function to be a primitive function. Therefore, it doesn't indicate whether it is a dx or dt function.
 

collide

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jannny said:
I somewat recall my teacher saying you can't take out t or x or y.. because how can u integrate if it is outside the Integral

Oh btw, I got this q from my the school holidays homework.. and I forgot to get the answers sheet.
Yeah I was thinking that as well.. but it just seemed like the most logical way to go about it :S
 

collide

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williamc said:
The notation in the original question showed the function to be a primitive function. Therefore, it doesn't indicate whether it is a dx or dt function.
Oh okay.. I just thought dt indicated that I was integrating with respect to t.
 
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pLuvia

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You should use partial fractions
Split the denominator into t(20-5t), then use partial fractions then integrate
 
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pLuvia

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It involves logarithmic methods after the partial fractions has been used
 

jb_nc

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f3nr15 said:
OLOL wheres your integral calculator ?
No, I did it myself using partial fractions. I wasn't going to type the 10 lines out and I double checked my answer in Matlab and the Integrator.

It's not that hard.
 
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