Question

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

Answer 1

`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)