cdf. expectation, and variance of a random continuous variable

Aug 2019
Given the probability density function f(x) = b[1-(4x/10-6/10)^2] for 1.5 < x <4. and f(x) = 0 elsewhere.

1. What is the value of b such that f(x) becomes a valid density function

2. What is the cumulative distribution function F(x) of f(x)

3. What is the Expectation of X, E[X]

4. What is the Variance of X, Var[X]

5. What is the probability that X is within one standard deviation from the mean

So far, I've gotten b by integrating the function from 1.5 to 4 and setting it equal to 1, thus getting b = 3/5.
Plugging b into the function, I also integrated it from 1.5 to x, to get a cdf of (-4x^3/125 + 18x^2/125 + 48x/125 - 99/125)
To get the E[X], I integrated f(x) from 1.5 to 4 to get 2.4375

However, I'm still kind of lost when it comes to the other 2 questions. I know that variance is supposed to be mean squared minus square of the mean. (E[X^2] - E[X]^2). However, plugging in x^2 instead of x and integrating from 1.5 to 4 gets me -158, which seems wrong as the variance would then be -164.
Further help would also be appreciated for number 5 as I am not even sure where to start on that one.