Find the solution to $\int_0^1 |2x-3y|dy$

Solution 1:

In general

$$ \int\vert u\vert\,du=\frac{1}{2}u\vert u\vert+c $$

which is easily verified from $\dfrac{d}{du}\vert u\vert=\dfrac{\vert u\vert}{u}=\dfrac{u}{\vert u\vert}$.

So

$$ \int\vert u\vert\,dy=\int\vert u\vert \frac{dy}{du}\,du $$

Let $u=2x-3y$, with $x$ constant. Then $\dfrac{dy}{du}=-\dfrac{1}{3}$ \begin{eqnarray} \int_0^1\vert 2x-3y\vert\,dy&=&-\frac{1}{3}\int_0^1\vert 2x-3y\vert\,du\\ &=&-\frac{1}{6}\left[(2x-3y)\vert2x-3y\vert\right]_0^1\\ &=&\frac{2}{3}x\vert x\vert-\frac{1}{6}(2x-3)\vert2x-3\vert \end{eqnarray}

This can be simplified for the three cases

1. $x<0$

2. $0\le x<\frac{3}{2}$

3. $x\ge\frac{3}{2}$

Here is a graph of the function: