Question

please help for this equations...http://www.webassign.net/cgi-bin/symimage.cgi?expr=%20int_0%5E%286%29%20abs%28x-3%29dx%20

Answer 1

You'll have to break the integral up into two parts.
\[\int_0^3 (3-x) \;dx + \int_3^6 (x-3)\;dx\]

Answer 2

so then its \[\int\limits_{0}^{6} (3-x)dx?\]

Answer 3

i meant = to that

Answer 4

one could also just find the areas of the two triangle (1/2*b*h) that the graph forms and add them together to get the answer of 9

Answer 5

You see, the absolute value cannot be integrated directly. If 0 < x < 3 then |x-3| = 3-x, and if 3<x<6 then | x-3| = x-3

Answer 6

yea Zarkon i was stuck doing that but couldnt see where i went wrong

Answer 7

so you have to treat the absolute value as two different functions, and integrate each part separately.

Answer 8

doing what cruffo suggests will lead to the correct answer too

Answer 9

i see what you are telling me Crruffo...but everone likes the short cut : )

Answer 10

yep... except some teachers :)