# P3 maths integration question

Watch
Announcements
This discussion is closed.
#1
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)
0
17 years ago
#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)
0
17 years ago
#3
What does the dollar sign mean? 0
#4
yeah sorry i meant dy not dx.

by the way the dollar sign means intergral. u know... the squigly line thing !
0
#5
(Original post by 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)
cheers elpaw that was great! u did it just like that! iwas expecting a few hours before someone got the answer for me. no wonder ur at oxford!. i bow down to ur greatness. 0
#6
however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
0
17 years ago
#7
(Original post by DLo)
however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
The original limits are wrt y. The new limits are wrt x.

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
0
X
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (330)
83.54%
No (65)
16.46%