I'm trying to find the probability of $P(\theta<0,10)$ knowing that $\theta $~Beta(2,20). So what I did was the integral:
$$\int_0^{0.10} \frac{\theta(1-\theta)^{19}}{B(2,20)}d\theta=3.16013\times 10^{-6}$$
However, following this website the probability is 0.64. Am I doing something wrong? Since the way I solved the integral was by changing the variables and nothing else.
Maybe you simply made some mistakes when doing the calculation.
$$\int_0^{0.1}\theta(1-\theta)^{19}d\theta = [-\frac{1}{20}\theta(1-\theta)^{20}]^{0.1}_0 + \frac{1}{20}\int_{0}^{0.1}(1-\theta)^{20}d\theta.$$
And by $t = 1-\theta$ we get
$$\int_{0}^{0.1}(1-\theta)^{20}d\theta = \int_{1}^{0.9}(t)^{20}(-dt) = \int_{0.9}^{1}t^{20}dt = [\frac{1}{21}t^{21}]^1_{0.9}\approx0.04240862$$.
$B(2,20)\approx 0.002380952$, $[-\frac{1}{20}\theta(1-\theta)^{20}]^{0.1}_0\approx -0.0006078833$. Combining them together gives the final answer 0.63527.