How to calculate probability of being with in one standard deviation of the mean value for a continuous distribution

Question

for a pdf of $f(x) = 90x^8(1-x)$, $0 < x < 1$, otherwise $f(x)=0$

I have calculated a of $F(x) = 10x^9 -9x^{10}$ for $0 < x < 1$

a mean of $9/11$

and a standard deviation of $\sqrt{.0123}$

I'm having a tough time figuring out what the $\mathbb P(A < x < B)$ would be, and if I use the integral of the pdf or the cdf to calculate said probability.

Answer 1

HINT

$$\mathbb P(A < x < B) = \mathbb P(x < B) - \mathbb P(x \le A) = F(B) - F(A)$$