Evaluate $$\int_0^1(1-2x)(x-x^2)^{2014} dx$$

Question

freckles · Answer

I was playing with @ganeshie8 's question and come across something weird....
I tried a substitution.
u=x-x^2
but this would make both upper and lower limit 0
But for some reason I'm not convinced the integral should evaluate to 0.

freckles · Answer

I was trying to look for a non-beta/gamma way.

freckles · Answer

http://openstudy.com/study#/updates/54a59652e4b054f0c3b68124 This was @ganeshie8 's post by the way.

turingtest · Answer

The graph of $f(x)=(1-2x)(x-x^2)^{2014}$ has 180 degree symmetry about x=1/2$$f(x+\frac12)=(1-2x-1)(x+\frac12-x^2-x-\frac14)=-2x(\frac14-x^2)$$In other words, it is "odd about x=1/2" (if I'm not just making up that terminology) hence any integral on an interval symmetric about x=1/2 will be zero.

turingtest · Answer

i forgot the exponent, but I doubt that causes confusion

solomonzelman · Answer

$\large\color{black}{\displaystyle\int\limits_{0}^{1}~(1-2x)(x-x^2)^{2014}dx}$

derivative of u, is sitting inside there, nice:)

solomonzelman · Answer

I mean if I set: u=x-x^2

solomonzelman · Answer

I will let other people do this, but nice integral, I like this:)

ganeshie8 · Answer

wana see some action movie

turingtest · Answer

haha yeah software doesn't like huge exponents

ganeshie8 · Answer

another way to look at TT's reply :)  :  
$$\int\limits_{0}^{1}(1-2x)(x-x^2)^{2014}dx $$

substitute $u = 1/2-x \implies 2u = 1-2x $ and $x-x^2 = x(1-x) = 1/4-u^2$

the integral becomes
$$\int\limits_{-1/2}^{1/2} (2u)(1/4-u^2)^{2014}\,du = 0$$

turingtest · Answer

that's what I wanted to write^, but I kept messing it up

ganeshie8 · Answer

guess we are just shifting the graph to left by 1/2 units so that it is symmetric about origin

freckles · Answer

ok yes I get it $$h(x)=x(x-x^2)^{2014}$$
is "odd about zero"
and $$f(x)=(1-2x)(x-x^2)^{2014}$$
is just a translation of the graph h
and it should be "odd about 1/2"
and nice picture

solomonzelman · Answer

don't get why can't we just set u=x-x^2, lol. it is made for that.

solomonzelman · Answer

it's pretty basic:)

solomonzelman · Answer

someone is trying to increase engagement score?  jk

turingtest · Answer

freckels had the answer, but realized that the substitution u-x-x^2 make the bounds both zero, and they wanted intuition as to why the integral should be zero.

solomonzelman · Answer

sure

turingtest · Answer

u=x-x^2
*

solomonzelman · Answer

I realize the typo:D