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where do x^4 and x^3 match up? 0 and 1 right?

try again:
2pi {S} x(x^4) - x(x^3) dx ; [0,1]

x^5 ints to x^6/6 and x^4 ints up to x^5/5

2pi/6 - 2pi/5 = ?

What's the {S}?

its the symbol I use to type in the integration elongated 's'

Ah, okay ^^
but doesn't 2pi/6 - 2pi/5 = -pi/15?

take the absolute value |-pi/15|

x^4 is flatter than x^3; so its actually below it
try .5^4 and .5^3 for trial

.5^4 is actually a smaller number than .5^3 i think

.625 and .125 right?

.0625 and .125 maybe

0.0625 and 0.125 the ^4 is less than the ^3

since we are working with numbers that are fractions; we get x^4 under x^3

anything greater than 1 reverses it

x^4 is under x^3 between 0 and 1; so it simply means you subtracted the lesser from the greater

and i proved that by:
(1/2)^4 < (1/2)^3
1/16 < 1/8

and when you subtract a lesser from a greater; it gives you a negative...

(3/4)^4 < (3/4)^3
81/256 < 27/64

x^3 is bigger then x^4 between 0 and 1

Ah! Okay, I get it now! ^^ Thanks so much!