Suppose that a random variable $\mathbb{X}$ has density $f$ and a unique median $m$ . Suppose that $b$ is any real number.
Show that $\mathbb{E(|X − b|) = E(|X − m|) + 2 \int_ b^ m (b − x)f(x)dx}$ , where we assume that the required expectations exist.
Let $m>b$. Then $$ \mathbb E(|X-b|) =$$ $$= \int\limits_{-\infty}^b bf(x)dx - \color{red}{\int\limits_{-\infty}^b xf(x)dx} +\int\limits_b^m xf(x)dx - \int\limits_b^m bf(x)dx +\color{blue}{\int\limits_m^\infty xf(x)dx} - \int\limits_m^\infty bf(x)dx $$ and $$ \mathbb E(|X-m|) =$$ $$=\! \int\limits_{-\infty}^b mf(x)dx -\! \color{red}{\int\limits_{-\infty}^b xf(x)dx} +\!\int\limits_b^m mf(x)dx -\! \int\limits_b^m xf(x)dx +\color{blue}{\int\limits_m^\infty xf(x)dx} -\! \int\limits_m^\infty mf(x)dx $$
Subtract the second expression from the first one. $$ \mathbb E(|X-b|)-\mathbb E(|X-m|) $$ $$ =(b-m)\int\limits_{-\infty}^b f(x)dx +\! 2 \int\limits_b^m xf(x)dx - \int\limits_b^m bf(x)dx -\! \int\limits_b^m mf(x)dx - \!(b-m)\int\limits_m^\infty f(x)dx $$
Then use that $\int\limits_m^\infty f(x)\,dx = \int\limits_{-\infty}^m f(x)\,dx$ since $m$ is the unique median. Replace last integral and subtract it from the first one: $$ \mathbb E(|X-b|)-\mathbb E(|X-m|) $$ $$ =\color{green}{(b-m)\int\limits_{-\infty}^b f(x)dx} + 2 \int\limits_b^m xf(x)dx - \int\limits_b^m bf(x)dx - \int\limits_b^m mf(x)dx - \color{green}{(b-m)\int\limits^m_{-\infty} f(x)dx} $$ $$ =(m-b)\int\limits_b^m f(x)dx+ 2 \int\limits_b^m xf(x)dx - \int\limits_b^m bf(x)dx - \int\limits_b^m mf(x)dx = 2\int\limits_b^m (x-b)f(x)\,dx. $$ Note also that $\mathbb E(|X-b|)$ should be greater than $\mathbb E(|X-m|)$. Therefore the integral in r.h.s. of the equality that you prove must be positive. So the right equality should be $$ \mathbb{E}(|X − b|) = \mathbb E(|X − m|) + 2 \int_b^m (x − b)f(x)dx. $$