How do you evaluate the definite integral #int abs(x-5) dx# from #[0,10]#?

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Answer 1 · 2016-09-08T10:24:51

25

#int_0^10 abs(x-5) \ dx#

#abs(x-5) = x-5, x in [5,oo)#

#abs(x-5) = -(x-5), x in [5,-oo)#

#implies int_0^5 -(x-5) \ dx + int_5^10 x-5 \ dx#

#= [ (x-5)^2/2 ]_5^0 + [ (x-5)^2/2 ]_5^10#

#= (25/2 ) + ( 25/2 ) = 25#

you can see this in the graph

graph{abs(x-5) [-5.375, 14.625, -2.16, 7.84]}