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Find the area between two curves (1 Viewer)

BlueGas

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The two curves are included in the picture below, I couldn't find the other x so it would be helpful if someone can help me but I want to also know if what I did so far in the picture was right, thanks.

 
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integral95

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multiply the whole equation by 3 to get





Divide by - 2 to get





therefore it's x = 3/2
 

BlueGas

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I continued the question today and this is my working out, my final answer is 1/12 (I didn't write it because it's wrong) but the real answer is 1/36. What did I do wrong?
 

jkerr138

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I have the solution of 1/36, but my photo is too big for this forum. Can I send the photo to you on an alternative address or medium?
 

BlueGas

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I have the solution of 1/36, but my photo is too big for this forum. Can I send the photo to you on an alternative address or medium?
If you can't post it then don't worry, but do you know where I went wrong?
 

jkerr138

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Yes, what you should have done:
1) Limits were wrong way round
2) First Integral should have been (2x-x^2) dx between x= 1.5 and x=1
3) That first integral should be deducted from the second integral (x^2/3 - 4/3 x +2) dx between x= 1.5 and x= 1
4) The answer you should get for part 2 is 11/24. The answer you should get for the second integral is 31/72.
5) Therefore as I said you must deduct the first integral from the second integral, you will have 11/24 - 31/72.
This will equal 1/36 units^2. The answer you were looking for. You needed to really inspect the graph to know what to do.
 

BlueGas

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Yes, what you should have done:
1) Limits were wrong way round
2) First Integral should have been (2x-x^2) dx between x= 1.5 and x=1
3) That first integral should be deducted from the second integral (x^2/3 - 4/3 x +2) dx between x= 1.5 and x= 1
4) The answer you should get for part 2 is 11/24. The answer you should get for the second integral is 31/72.
5) Therefore as I said you must deduct the first integral from the second integral, you will have 11/24 - 31/72.
This will equal 1/36 units^2. The answer you were looking for. You needed to really inspect the graph to know what to do.
Ahh that's what I did wrong, I thought 1 has higher than 3/2, thanks alot man.
 

BlueGas

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Yes, what you should have done:
1) Limits were wrong way round
2) First Integral should have been (2x-x^2) dx between x= 1.5 and x=1
3) That first integral should be deducted from the second integral (x^2/3 - 4/3 x +2) dx between x= 1.5 and x= 1
4) The answer you should get for part 2 is 11/24. The answer you should get for the second integral is 31/72.
5) Therefore as I said you must deduct the first integral from the second integral, you will have 11/24 - 31/72.
This will equal 1/36 units^2. The answer you were looking for. You needed to really inspect the graph to know what to do.
How come for this question I didn't need two integrals? I only used one as you can see in the picture, that's kind of confusing me.



EDIT: I'm speaking about question 5 by the way.
 
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