Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (caozeyuan):

question about reduction formula posted below. Anyone with knowledge on calculus are welcomed!

OpenStudy (caozeyuan):

Let \[I _{n}= \int\limits_{0}^{1} x ^{n}\left( 1-x \right)^{\frac{ 1 }{ 2}}dx\]

OpenStudy (caozeyuan):

Show that, \[n \ge 1, \left( 3+2n \right)I _{n}=2nI _{n-1}\]

OpenStudy (caozeyuan):

Hence found the exact value of I subscript 3

OpenStudy (caozeyuan):

@amistre64

OpenStudy (caozeyuan):

@terenzreignz

terenzreignz (terenzreignz):

Where have I 'seen' you before...? name looks familiar

terenzreignz (terenzreignz):

Regardless... I know next-to-nothing about reduction formulas... however, let's have a look see at this :D

OpenStudy (caozeyuan):

If you are living in Beijing, then you might know me personally

terenzreignz (terenzreignz):

Unlikely... besides, my grandparents are (originally) from Guangzhou XD

terenzreignz (terenzreignz):

Back to the question...

OpenStudy (caozeyuan):

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20-%20Further%20(9231)/9231_w11_ms_11.pdf the official answer is given in this document, page 7, question 6. But it is overly simplified

terenzreignz (terenzreignz):

^That's like a spoiler... I will resist clicking on that link :D

OpenStudy (caozeyuan):

@Loser66 , Aha! another great mathematician comes online!

OpenStudy (caozeyuan):

Are you serious, or just kidding

terenzreignz (terenzreignz):

About what?

OpenStudy (caozeyuan):

Never mind, I've figured it out from the page, suggesting you have a look. Here is the link to the question paper http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20-%20Further%20(9231)/9231_w11_qp_11.pdf

terenzreignz (terenzreignz):

And to think I was having a real good think XD

terenzreignz (terenzreignz):

Although the solution seems very straightforward.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!