2011 past paper question (1 Viewer)

[luvglee4lyfe]

Hey Bos'ers! I'm new here and i just want some maths help. this questino is from the 2011 hsc paper i think.

I need help with the second part. I know how to do the first part.

 

[nerdasdasd]

Hey Bos'ers! I'm new here and i just want some maths help. this questino is from the 2011 hsc paper i think.

I need help with the second part. I know how to do the first part.

Hint: try express the function in a different form. E.g. y = a^2.

 

[btx3]

integration is stoopid, the sooner you learn that, the better

 

[Parvee]

try rearranging the integral in the form of -x/[sqrt(9-x^2)] i.e. your answer from i)

You should get your answer from this

 

[albertcamus]

the derivative of first part is (-x/sqrt(9-x^2))

to get 6x from that you multiple -x by -6

.'. it becomes -6∫(-x/sqrt(9-x^2))

and that's just -6 x the primitive of (-x/sqrt(9-x^2))

which is -6 x sqrt 9-x^2

 

[nerdasdasd]

y = (9 - x) ^ 1/2

 

[luvglee4lyfe]

Hint: try express the function in a different form. E.g. y = a^2.

with the second part? i did that for the first part. i got the answer out i just need confirmation. =/

 

[luvglee4lyfe]

the derivative of first part is (-x/sqrt(9-x^2))

to get 6x from that you multiple -x by -6

.'. it becomes -6∫(-x/sqrt(9-x^2))

and that's just -6 x the primitive of (-x/sqrt(9-x^2))

which is -6 x sqrt 9-x^2

THANK YOU!! thats what i got too!

 

[Carrotsticks]

Hint: try express the function in a different form. E.g. y = a^2.

y = (9 - x) ^ 1/2

This only works if the inside of the bracket is linear.

The inside in this case is x^2, so we cannot use that method.