How do you evaluate the definite integral #int (3-abs(x-3))dx# from [1,4]?
1 Answer
Nov 10, 2016
Explanation:
The 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
#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#
