Find the area between two curves (1 Viewer)

[BlueGas]

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.

 

[BLIT2014]

You forgot to include a picture

 

[BlueGas]

You forgot to include a picture

I didn't realise, lol, just included the photo now.

 

[integral95]

multiply the whole equation by 3 to get





Divide by - 2 to get





therefore it's x = 3/2

 

[BlueGas]

multiply the whole equation by 3 to get





Divide by - 2 to get





therefore it's x = 3/2

Which equation do I multiply by 3?

 

[integral95]

Which equation do I multiply by 3?

your very last one

At the bottom of your whiteboard

 

[BlueGas]

your very last one

At the bottom of your whiteboard

So basically you just multiplied by three to get rid of the fractions, I should of knew that, thanks man.

 

[BlueGas]

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]

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]

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]

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]

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]

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.