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~ Yet another intigration question - HSN forum

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~ Yet another intigration question


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#1 Sammie

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Posted 14 January 2007 - 03:29 PM

Ok here we go ~

Part one.

Find the area bound by the curve Y = 6[root]x , the x-axis and the line x = 4


So for this part I have


Y = 6[root]X
Y = 6X1/2

= integral.gif 40 (6X1/2) DX
= (( 6X1/2) / (3/2))
= (4(4)1/2)
= 32 units2

Part Two.

The line X = a divides this area in half. Find a correct to 1 Decimal place.


Area = 32 units2
=> Area enclosed by Y = a
=> 16 units2

~~~ Then I get stuck...I'm just not too sure how to approach this...
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#2 KhalidBoussouara

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Posted 14 January 2007 - 04:09 PM

QUOTE(Sammie @ Jan 14 2007, 03:29 PM) View Post

Ok here we go ~

Part one.

Find the area bound by the curve Y = 6[root]x , the x-axis and the line x = 4


So for this part I have


Y = 6[root]X
Y = 6X1/2

= integral.gif 40 (6X1/2) DX
= (( 6X1/2) / (3/2))
= (4(4)1/2)
= 32 units2

Part Two.

The line X = a divides this area in half. Find a correct to 1 Decimal place.


Area = 32 units2
=> Area enclosed by Y = a
=> 16 units2

~~~ Then I get stuck...I'm just not too sure how to approach this...


The first bit is correct but it should be written as 4*43/2 and not 4*41/2

For the second bit your basically working backwards. The limits would be a and 0.

16 = integral.gif a0 (6X1/2) dx
= [( 6x3/2) / (3/2)]
= 4 a 3/2
4 = a 3/2

And I'm not sure how to type the next bit. But a would be the cubed root of 42.

The cubed root of 16 is 2.5 to one decimal place. So you've got a.
MSN: khalid@boussouara.co.uk





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