Question

does anyone know the meaning of this integration? what are the extra signs {,min, etc for? please help!
http://i46.tinypic.com/241qzw0.jpg

Answer 1

@phi @KonradZuse

Answer 2

\[\int\limits_{x}^{x^2}dx\] imo

Answer 3

but I could be way wrong, never seen it that way....

Answer 4

It's supposed to mean this:\[\dfrac{x + x^2 - \sqrt{x - x^2}}{2}\]

Answer 5

The \(\min\{x,x^2\}\) is equal to the above expression

Answer 6

Am I right?

Answer 7

min{x,x^2} dx
so when is x > x^2 ?
and when is x < x^2 ?
to know, take x=x^2, so, between 0 and 1, x>x^2
and when x>1, x^2 > x
so, your integral will split into 2 integrals,
\(\int_0^1 x^2dx +\int_1^{\infty}x dx\)

Answer 8

Ah.

Answer 9

but even i am not sure, because here the 2nd integral, does not converge...

Answer 10

my prof was trying to solve it on the board and this is how he started: i don't understand a thing!

Answer 11

@hartnn

Answer 12

I would assume it is the minimum of (x, x^2)
in other words, as a function of what interval you are in, either x or x^2 will be the minimum function.

for the interval -inf to 0, x is the min
for the interval 0 to 1, x^2 is the min (plot it, and you see x^2 < x between 0 and 1)
and for the interval 1 to +inf, x is the min

so you divide your integral into 3 intervals

Answer 13

I think you do this
\[\int\limits_{-\infty}^{0}x dx+\int\limits_{0}^{1}x^2 dx+\int\limits_{1}^{+\infty}x dx\]
now re-write the first integral by switching the order of the integrals
\[- \int\limits_{0}^{\infty}x dx+\int\limits_{0}^{1}x^2 dx+\int\limits_{1}^{+\infty}x dx\]
or
\[\int\limits_{0}^{1}x^2 dx+\int\limits_{1}^{+\infty}x dx- \int\limits_{1}^{\infty}x dx- \int\limits_{0}^{1}x dx\]
and finally
\[\int\limits_{0}^{1}x^2 -x dx\]