Integration

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Anonymous1502
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#1
Report Thread starter 4 years ago
#1
I have this question:

∫upper limit 2a lower limit a (10-6x)dx=1 find two possible values of a

I first integrated 10-6x which would give me 10x-3x^2 then I inserted the lower and upper limit and took them away

10(2a)-2(2a)^2=20a-12a^2
(20a-24a^2)-(10a-3a^2)=10a-9a^2

the mark scheme tells me to rearrange equation to form quadratic equation of

9a^2-10a+1

My question is why is it 1?I know it says that this is all equal to 1 but how do we know that c is 1?
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username3509752
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#2
Report 4 years ago
#2
10a - 9a^2 = 1 -->
0 = 1 - 10a + 9a^2
or 9a^2 - 1a + 1 = 0.
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Anonymous1502
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#3
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#3
(Original post by Edgemaster)
10a - 9a^2 = 1 -->
0 = 1 - 10a + 9a^2
or 9a^2 - 1a + 1 = 0.
Thank you very much.
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Anonymous12120
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#4
Report 2 years ago
#4
Hi I'm currently doing the same question. Where can you find the mark scheme for it?
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Anonymous2736
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#5
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#5
You probably don’t care about this anymore but it is one as you know the integration is 1 because it is stated in the question (∫upper limit 2a lower limit a (10-6x)dx=1). Therefore the integration of 20a-9a^2=1. This can then be rearranged to 9a^2-10a 1=0
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CatInTheCorner
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#6
Report 3 months ago
#6
(Original post by Anonymous2736)
You probably don’t care about this anymore but it is one as you know the integration is 1 because it is stated in the question (∫upper limit 2a lower limit a (10-6x)dx=1). Therefore the integration of 20a-9a^2=1. This can then be rearranged to 9a^2-10a 1=0
My man's going to travel back a year in time
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Anonymous2736
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#7
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#7
(Original post by Anonymous1502)
I have this question:

∫upper limit 2a lower limit a (10-6x)dx=1 find two possible values of a

I first integrated 10-6x which would give me 10x-3x^2 then I inserted the lower and upper limit and took them away

10(2a)-2(2a)^2=20a-12a^2
(20a-24a^2)-(10a-3a^2)=10a-9a^2

the mark scheme tells me to rearrange equation to form quadratic equation of

9a^2-10a+1

My question is why is it 1?I know it says that this is all equal to 1 but how do we know that c is 1?
You probably don’t care about this anymore but for anyone looking at this question later on… it is 1 because, as the question states, the integration is equal to 1 (∫upper limit 2a lower limit a (10-6x)dx=1). So the integration you have found of 10a-9a^2 equals 1 (10a-9a^2=1). From this you can rearrange to get 9a^2-10a+1=0.
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Anonymous2736
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#8
Report 3 months ago
#8
(Original post by CatInTheCorner)
My man's going to travel back a year in time
Eh, it was mostly in case anyone else searched for the question. I got it last week and was confused by it as well. Figured I could try and explain it the way I understood it.
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