Question

Beta functions
Please help :)

Answer 1

\(\Huge\int_{0}^{2a}x^2\sqrt{(2ax-x^2)}dx\)

Answer 2

here
\[\int\limits_{0}^{2a}\sqrt x~\cdot~x^2 \sqrt{(2a-x)} dx \]
let x=2at
dx=2adt

Answer 3

correct?

Answer 4

Looks perfect!

Answer 5

i get
\(\rm 2^{1/2}~\cdot~8\sqrt2~\cdot~a^4~\cdot~\beta(7/2,3/2) \)

Answer 6

\(5\pi/8\)

Answer 7

correct?

Answer 8

http://www.wolframalpha.com/input/?i=beta%287%2F2,+3%2F2%29

Answer 9

correct!

Answer 10

Thank you once again
i have one more question should i post it here ?

Answer 11

Sure...

Answer 12

thanks !

\[\int\limits_{0}^{1}\frac{ x^{3/2} }{ \sqrt{3-x} }dx \cdot \int\limits_{0}^{1}\frac{ dx }{ \sqrt{1-x^{1/4}}}\]

Answer 13

it is a single question :)

Answer 14

i m not getting the substitution

Answer 15

are you sure the bounds are correct ?

Answer 16

it is the same given in the question
i suppose it maybe wrong in the question

Answer 17

\[ \int\limits_{0}^{1}\frac{ dx }{ \sqrt{1-x^{1/4}}}\]

this can be easily expressed in terms of beta integral by substituting \(x^{1/4}=t\)

Answer 18

yup but the first part?

Answer 19

yeah that is where i was stuck

the bounds are getting changed if we let x = 3t

Answer 20

yup i took the same
but :(

Answer 21

it would be easy if the bounds were 0 to 1/3

Answer 22

yup

Answer 23

umm
ill check that question later with others

see this one :

\[\rm I=\int\limits_{0}^{1}\frac{ dx }{ \sqrt{(1-x^6)} }\\ Let~x=t^{1/6}~~\\ I= \int\limits_{0}^{1}\frac{ 1 }{6 }\cdot~t^{(1/6)-1}\cdot(1-t)^{1/2-1}dt~~\\I=\frac{ 1 }{ 6 }~\beta(1/6,1/2)\]

Answer 24

Looks good!

Answer 25

sorry it was lagging! :(

Answer 26

so yes now again that beta function @ganeshie8