# P3 maths integration question

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hey guys. i was wondering is all the good mathematicians out there could help me on a hard question. (if it helsp this question is question 7 on an OCR past P3 paper dated 16 January 2002). i dont have a clue on thsi so could someone go through it step by sytep.

by using substitution y=a sinx to prove that the

$ (limits a and a/2) (a² - y²)^½ dx = (a²/2)[(n/3) - (3^½)/2]

$ means integral

^½ means to the power of a half

n means pi (i.e. 3.14)

by using substitution y=a sinx to prove that the

$ (limits a and a/2) (a² - y²)^½ dx = (a²/2)[(n/3) - (3^½)/2]

$ means integral

^½ means to the power of a half

n means pi (i.e. 3.14)

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#2

you've written dx, don't yoou mean dy?

y = a sinx => dy/dx = a cosx

$ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= $ (limits pi/6 and pi/2) a² cos²x dx

= $ (limits pi/6 and pi/2) a²/2 (1+cos2x) dx

=a²/2[x + 1/2 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)

y = a sinx => dy/dx = a cosx

$ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= $ (limits pi/6 and pi/2) a² cos²x dx

= $ (limits pi/6 and pi/2) a²/2 (1+cos2x) dx

=a²/2[x + 1/2 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)

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yeah sorry i meant dy not dx.

by the way the dollar sign means intergral. u know... the squigly line thing !

by the way the dollar sign means intergral. u know... the squigly line thing !

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(Original post by

you've written dx, don't yoou mean dy?

y = a sinx => dy/dx = a cosx

$ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= $ (limits pi/6 and pi/2) a² cos²x dx

= $ (limits pi/6 and pi/2) a/2 (1+cos2x) dx

=[ax/2 + a/4 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)

**elpaw**)you've written dx, don't yoou mean dy?

y = a sinx => dy/dx = a cosx

$ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

= $ (limits pi/6 and pi/2) a² cos²x dx

= $ (limits pi/6 and pi/2) a/2 (1+cos2x) dx

=[ax/2 + a/4 sin2x] (limits pi/6 and pi/2)

= a²/2(pi/3 - root3/2)

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however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?

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#7

(Original post by

however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?

**DLo**)however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?

using the substitution y=asinx,

at y=a/2, a/2=asinx => sinx = ½ => x=pi/6

at y=a, a=asinx => sinx = 1 => x=pi/2

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