How do you evaluate the definite integral #int (3-abs(x-3))dx# from [1,4]?

1 Answer
Nov 10, 2016

#int_1^4(3-abs(x-3))dx=13/2#

Explanation:

The function #abs(x-3)# can be expressed as a piecewise function:

#abs(x-3)={(x-3,,x>=3),(-(x-3),,x<3):}#

So, we can write our integral as the sum of two integrals, breaking off at #x=3#:

#int_1^4(3-abs(x-3))dx#

#=int_1^3(3-(-(x-3)))dx+int_3^4(3-(x-3))dx#

#=int_1^3(3+(x-3))dx+int_3^4(3-x+3)dx#

#=int_1^3xcolor(white).dx+int_3^4(-x+6)dx#

#=[x^2/2]_1^3+[-x^2/2+6x]_3^4#

#=(3^2/2-1^2/2)+(-4^2/2+6(4))-(-3^2/2+3(6))#

#=(9/2-1/2)+(-8+24)-(-9/2+18)#

#=13/2#