Median for continuous distribution 
Consider a continuous random variable X with probability density function given by $f(x)=cx$ for $1 \le x \le 5$, zero otherwise.  Find the median.

First I calculate the CDF:  $F(x)=cx^2/2$ for $1 \le x \le 5$, zero otherwise.
Now we have to solve for constant c by using the definition of PDF, namely:
$$\int\limits_{-\infty}^{\infty}f(x)dx=1 \implies \frac{c}{2}x^2\Big{|}_1^5=1 \implies c=\frac{1}{12} $$
Then to calculate the median, we set the CDF = 0.5:
$$\frac{1}{2}=\frac{1}{12}\cdot \frac{1}{2} \cdot x^2 \implies x=\sqrt{12}$$
But the book solution is $\sqrt{13}$.  Can someone tell me what I am doing wrong?
Thank you.

For median $m$:
$$
\int\limits_{-\infty}^{m} f(x) dx = \int\limits_{m}^{+\infty} f(x) dx\\
\int\limits_{1}^{m} f(x) dx = \int\limits_{m}^{5} f(x) dx\\
c \frac{x^2}{2}\Biggr|_{1}^{m} = c \frac{x^2}{2}\Biggr|_{m}^{5}\\
x^2\Biggr|_{1}^{m} = x^2\Biggr|_{m}^{5}\\
m = \sqrt{\frac{1^2 + 5^2}{2}}=\sqrt{13}
$$
