1 reply to this topic

[Sammie]

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 = 6X^1/2

= ⁴₀ (6X^1/2) DX
= (( 6X^1/2) / (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...

[KhalidBoussouara]

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 = 6X^1/2

= ⁴₀ (6X^1/2) DX
= (( 6X^1/2) / (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*4^3/2 and not 4*4^1/2

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

16 = ^a₀ (6X^1/2) dx
= [( 6x^3/2) / (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.