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Integration

How do you integrate when it is a fraction with the unknown value being the denominator. For example, how would you integrate 12/x^2?

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Reply 1
Rearrange it so that you get 12x^-2 then just do what you do… i was just about to say what happens if it's like 12/x with no power but that won't come up at GCSE
(edited 7 years ago)
Reply 2
Original post by GCSE2016Troop
How do you integrate when it is a fraction with the unknown value being the denominator. For example, how would you integrate 12/x^2?


Note that 1x2=x2\frac{1}{x^2} = x^{-2} so 12x2dx=12x2dx\int \frac{12}{x^2} \, \mathrm{d}x = \int 12x^{-2} \, \mathrm{d}x and I trust you know how to integrate a power of xx?
Original post by GCSE2016Troop
How do you integrate when it is a fraction with the unknown value being the denominator. For example, how would you integrate 12/x^2?


Use laws of indices to write 12x2=12x2 \frac{12}{x^2} = 12x^{-2}


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Original post by Zacken
Note that 1x2=x2\frac{1}{x^2} = x^{-2} so 12x2dx=12x2dx\int \frac{12}{x^2} \, \mathrm{d}x = \int 12x^{-2} \, \mathrm{d}x and I trust you know how to integrate a power of xx?


lol id hope so thanks
Reply 5
Original post by zayn008
Rearrange it so that you get 12x^-2 then just do what you do… i was just about to say what happens if it's like 12/x with no power but that won't come up at GCSE


12xdx=12logx+C\int \frac{12}{x} \, \mathrm{d}x = 12\log x + \mathrm{C}. That's a good question, it's a shame that not enough students ask themselves it.
Reply 6
Original post by Zacken
12xdx=12logx+C\int \frac{12}{x} \, \mathrm{d}x = 12\log x + \mathrm{C}. That's a good question, it's a shame that not enough students ask themselves it.


12lnx+C 12\ln x + \mathrm{C}. * :tongue:
Reply 7
Original post by zayn008
12lnx+C 12\ln x + \mathrm{C}. * :tongue:


logx=lnx\log x = \ln x for mathematicians.
Reply 8
Original post by Zacken
logx=lnx\log x = \ln x for mathematicians.


Oh, sorry :colondollar:
Original post by Zacken
Note that 1x2=x2\frac{1}{x^2} = x^{-2} so 12x2dx=12x2dx\int \frac{12}{x^2} \, \mathrm{d}x = \int 12x^{-2} \, \mathrm{d}x and I trust you know how to integrate a power of xx?


Hi please can you help me with this question. You had to integrate 1 + 12/x^2 + 48/x^4 + 64/x^6. From this I got -4x^-3 + -48/5x^-5 + -64/7x^-7. You then had to integrate between the values of 2 and 1. It was pretty long so I'm just going to write out my whole working but in the end I got 1531/70. I feel like I've gone wrong somewhere along away possibly in my initial integration or when plugging the values in I don't know please could you help thanks.
Reply 10
Original post by GCSE2016Troop
Hi please can you help me with this question. You had to integrate 1 + 12/x^2 + 48/x^4 + 64/x^6. From this I got -4x^-3 + -48/5x^-5 + -64/7x^-7. You then had to integrate between the values of 2 and 1. It was pretty long so I'm just going to write out my whole working but in the end I got 1531/70. I feel like I've gone wrong somewhere along away possibly in my initial integration or when plugging the values in I don't know please could you help thanks.


you forgot to integrate 1.
Original post by Zacken
you forgot to integrate 1.


god sakeeeeeeeeeeeeeeeeeeeeee
Original post by Zacken
you forgot to integrate 1.


Is the other parts of the integration correct?
Reply 13
Original post by GCSE2016Troop
Is the other parts of the integration correct?


yeah, your final answer should be 167/5 if you wanted verification
Original post by Zacken
yeah, your final answer should be 167/5 if you wanted verification


How did you get that? I keep getting 79/70 for the first value and -1277/70 for the second value and so got 1356/70 overall?
....
(edited 7 years ago)
Original post by Zacken
logx=lnx\log x = \ln x for mathematicians.


Sorry just putting 'log' implies log to the base 10 and that is wrong.
Original post by zayn008
Oh, sorry :colondollar:


You are correct, Zacken is wrong.
Reply 18
Original post by GCSE2016Troop
How did you get that? I keep getting 79/70 for the first value and -1277/70 for the second value and so got 1356/70 overall?


Uh, my bad - your integration isn't correct. 12x2=12+1x2+1=12x\int 12x^{-2} = \frac{1}{-2 + 1} x^{-2 + 1} = \frac{-12}{x}, same for all your other terms. I think you're getting confused with differentiation.
Reply 19
Original post by Muttley79
Sorry just putting 'log' implies log to the base 10 and that is wrong.


A great many of my lecturers are wrong then. :beard:

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