Integration

This is a Putnam question - Putnam being an (advanced) American university student mathematics competition. I'm finding this integration question a pain in the ass. I don't really need the solution, but would like to see it if anyone gets it.

Take a look at it, and have fun.

ignore this

I get 1.

lol using a calculator / computer?

Reply 4

ryan750

2

JFN

This is a Putnam question - Putnam being an (advanced) American university student mathematics competition. I'm finding this integration question a pain in the ass. I don't really need the solution, but would like to see it if anyone gets it.

Take a look at it, and have fun.

could u show me how to do it?

Reply 5

Newton

It is impossible to find the integral of this function. Hence we use an estimate prediction achieved by the Trapezium Rule. Let us say with 5 strips of equal width:

h=(4-2)/(5)=0.4

f(2.0)=0.52

f(2.4)=0.52

f(2.8)=0.51

f(3.2)=0.51

f(3.6)=0.50

f(4.0)=0.49

By using the "Trapezium Rule":

A=1.018 (VERY CLOSE to 1.0)

Depending on the weighing of this question if you use 10 strips, your answer would get closer to 1.0 and so forth. Therefore:

A=1(units^2) (to 1d. p.)

Newton.

Reply 6

JamesF

1

Ummm, I doubt that its impossible. Plug it into the integrator and you get a very simple expression for the indefinite integral. You just gotta work out how to simplify it.

Reply 7

JamesF

1

Oh yea, forgot...there will be a solution on here somewhere

Reply 8

JFN

OP

1

Its not impossible. Its simply complicated - thats the way the exam is. I've attached another question - incidentally on integration too. Its also quite difficult..

Question 1 is beautiful - the answer is exactly 1.
Will have to think abt Q2.

Davros

Reply 10

Jonny W

8

For the first problem, try substituting (1) x = 3 + y, and (2) x = 3 - y. Then show that the sum of the two resulting integrals is 2.

Rep for the first person who finds numbers A and B such that

(int from A to B) f(2004 - x) / [f(2004 - x) + f(x + 1066)] dx = 1

for any function f.

Reply 11

davros

16

I did Q1 with just 1 substitution: u = 6 - x which preserves the limits of 2 and 4; then just add the new integral back to the old one (getting 2I = 2 where I is the value of the integral).

Davros

Reply 12

evariste_3086

1

Jonny W

For the first problem, try substituting (1) x = 3 + y, and (2) x = 3 - y. Then show that the sum of the two resulting integrals is 2.

Rep for the first person who finds numbers A and B such that

(int from A to B) f(2004 - x) / [f(2004 - x) + f(x + 1066)] dx = 1

for any function f.

let I=(int from A to B) f(2004 - x) / [f(2004 - x) + f(x + 1066)] dx
with A=468 B=470
since
let 2004-y=x+1066 so dy=-dx
when x=468 y=470
when x=470 x=468
so I=(int 468 to 470)f(y+1066)/[f(2004 - y) + f(y + 1066)] dy
so 2I=integral of 1 between 470 and 468=2
so I=1

Reply 13

davros

16

I agree with Evariste - I got 468 and 470 too.

Anyone had a look at the 2nd question - haven't really had a serious attempt myself...

Davros

Integration

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