\begin{equation}
\begin{split}
\int_K x \,dx &=\int_{K\cap [0,m(K)]} x\,dx + \int_{K\cap [m(K), d]} x \,dx\\
&\ge \int_{K\cap [0,m(K)]} x\,dx + m(K) m(K \cap [m(K), d]) \ \ \ \ \ \ \ (1)\\
&= \int_{K\cap [0,m(K)]} x\,dx + m(K) m([0, m(K)] \setminus K)\ \ \ \ \ \ \ (2) \\
&\ge \int_{K\cap [0,m(K)]} x\,dx + \int_{[0, m(K)]\setminus K} x \,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
&= \int_0^{m(K)} x\,dx.
\end{split}
\end{equation}
Comentario
(1):$x\ge m(K)$$K\cap [m(K), d]$,
$$ \int_{K\cap [m(K), d]} x \,dx \ge \int_{K\cap [m(K), d]} m(K) \,dx = m(K) m(K\cap [m(K), d])$$
(2) tenga en cuenta que $K \subset [0,d] = [0,m(K)] \cup [m(K), d]$, por lo que
$$m(K) = m(K \cap [0, m(K)]) + m(K \cap [m(K), d])$$
Por otro lado,
$$m(K) = m([0,m(K)]) = m(K\cap [0, m(K)]) + m([0,m(K)]\setminus K)$$
Estas dos igualdades implica
$$m(K \cap [m(K), d]) = m([0,m(K)]\setminus K).$$
(3) Similar (1), como $x \le m(K)$$[0,m(K)]$, tenemos
$$\int_{[0, m(K)]\setminus K} x \,dx \le \int_{[0, m(K)]\setminus K} m(K) \,dx = m(K) m([0,m(K)]\setminus K)$$