Question

An interesting and challenging but very basic problem

Answer 1

If \(\large{A = \int\limits_{0}^{1}\prod_{r=1}^{2014}(r-x)dx}\) and \(\large{B = \int\limits_{0}^{1}\prod_{r=0}^{2013}(r+x)dx}\), then the relation between A and B is

(a) A = 2B
(b) 2A = B
(c) A+B = 0
(d) A = B

Answer 2

@ganeshie8 , @Miracrown , @dan815

Answer 3

Does anyone want a hint ?

Answer 4

No.. I'm trying.

Answer 5

Ok take ur time

Answer 6

@vishweshshrimali5 tell me if I'm on the right track... Would you write the product as the sum of logarithms?

Answer 7

well you may use that but its going to make the problem more complex. Try something more basic.

Answer 8

A = pi (r = 1 to 2014) ( r - 1/2) ...

does this make any sense?

Answer 9

Do not forget integration and its wrong.

Answer 10

Thanks! :P

Answer 11

Hint ??

Answer 12

You should give us a few hours before you do that :)

Answer 13

No problem.

Answer 14

Just one thing to all who are trying to solve this question: Do not judge a book by its cover and never try to judge the complexity of a question just by looking at its symbols.

Answer 15

A + B = 0

Answer 16

no

Answer 17

o.O That's unfair.. I did a lot of hard work to get to that... ! :(

Answer 18

Sorry

Answer 19

;)

Answer 20

Its okay :P

Answer 21

I want to say \(A=B\)...

Answer 22

Good but how ?

Answer 23

\(\large{\int\limits_{0}^{1}\prod_{r=0}^{2013}(r-x)dx - 1 = B}\)

Answer 24

A shift in the index, I think. Something like \(s=r+1\).

Answer 25

Well now let me give the solution

Answer 26

I did it above :P

Answer 27

Solve that.. don't be lazy!

Answer 28

\[\large{A = \int\limits_{0}^{1}\prod_{r=1}^{2014}(r-x)dx}\]

\[\large{\implies A = \int\limits_{0}^{1}(1-x)(2-x)(3-x)...(2014-x)dx}\]

\[\large{\implies A = \int\limits_{0}^{1}[1-(1-x)][2-(1-x)][3-(1-x)]...[2014-(1-x)]dx}\]

\[\large{\implies A =\int\limits_{0}^{1}(x)(x+1)(x+2)...(x+2013)dx}\]

\[\large{\implies A = \int\limits_{0}^{1}\prod_{r=0}^{2013}(r+x)dx}\]

\[\large{\implies A = B}\]

Answer 29

See how basic but interesting !!

Answer 30

you put x = 1-x ?
o.O

Answer 31

I applied this property here:

\[\large{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(b+a-x)dx}\]

Answer 32

You must have told the property to us first... then we could have solved it :P

But, that's a great thing to learn. Awesome question and awesome technique @vishweshshrimali5 ! Great.

Answer 33

Hey, I resent it ;)

And thanks

Answer 34

;)

Answer 35

Well lets give you another integration problem.