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

Nov 29, 2016

$\int f \left(t\right) \mathrm{dt} = | {t}^{2} - 4 t | + c$

#### Explanation:

Let $f \left(t\right) = | 2 t - 4 |$

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

Then we can write;

$f \left(t\right) \setminus \setminus \setminus \setminus \setminus \setminus = \left\{\begin{matrix}- \left(2 t - 4\right) & t < 0 \\ 0 & t = 0 \\ 2 t - 4 & t > 0\end{matrix}\right.$

$\therefore f \left(t\right) \setminus = \left\{\begin{matrix}- 2 t + 4 \setminus \setminus \setminus & t < 0 \\ 0 & t = 0 \\ 2 t - 4 & t > 0\end{matrix}\right.$

So then it should be obvious that:

$\therefore \int f \left(t\right) \mathrm{dt} = \left\{\begin{matrix}- {t}^{2} + 4 t + {c}_{1} & t < 0 \\ {c}_{2} & t = 0 \\ {t}^{2} - 4 t + {c}_{3} & t > 0\end{matrix}\right.$

Where ${c}_{1} , {c}_{2} , {c}_{3}$ are arbitrary constants

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

$\therefore \int f \left(t\right) \mathrm{dt} = \left\{\begin{matrix}- \left({t}^{2} - 4 t\right) + {c}_{1} & t < 0 \\ {t}^{2} - 4 t + {c}_{2} & t = 0 \\ | {t}^{2} - 4 t | + {c}_{3} & t > 0\end{matrix}\right.$

Hence we can write
$\int f \left(t\right) \mathrm{dt} = | {t}^{2} - 4 t | + c$
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)