How do you find the antiderivative of #abs(2t-4)#?

1 Answer
Nov 29, 2016

# int f(t) dt= |t^2-4t| + c#

Explanation:

Let # f(t) = |2t-4| #

graph{|2x-4| [-9.75, 10.25, -1.44, 8.56]}

Then we can write;

# f(t) \ \ \ \ \ \ = { (-(2t-4), t<0), (0, t=0), (2t-4, t>0) :} #

# :. f(t) \ = { (-2t+4\ \ \ , t<0), (0, t=0), (2t-4, t>0) :} #

So then it should be obvious that:

# :. int f(t) dt = { (-t^2+4t + c_1 , t<0), (c_2, t=0), (t^2-4t+c_3, t>0) :} #

Where #c_1, c_2, c_3# are arbitrary constants

With a bit of manipulation you should see that we can write

# :. int f(t) dt = { (-(t^2-4t)+c_1 , t<0), (t^2-4t+c_2, t=0), (|t^2-4t|+c_3, t>0) :} #

Hence we can write
# int f(t) dt= |t^2-4t| + c#
as a solution (but not the general solution, as the +ve and -ve and zero sections of the solution could have a different constant of integration)