Prove the median of a uniform distributions is $\frac{1}{2}(a+b)$

Question

Let X~U(a,b) with a and b in the real line, such that b>a with X's pdf given by $$f_X(x)=\frac{1}{b-a}\mathbb{1}(a<x< b)$$ Show the median of X's distribution is given by $$m=\frac{1}{2}(b+a)$$ So I started by saying $$\frac{1}{2} = \int_{0}^{m}\frac{1}{b-a}dx$$ but when I work this out I get $$m=\frac{1}{2}(b-a)$$ Does anyone know why it changes from (b-a) to (b+a)?

Answer 1

The integral $\int_a^m \frac{1}{b-a}dx$ represents the cumulative density function for the uniform distribution. Median is point where the density function is $1/2$. If you change your lower limit to $a$, you should see this happening. For the uniform distribution, median is same as the expected value.