Finding probability interval from cummulative distribution function

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I've been asked to find the cumulative distribution function of :

$$f(x) = \begin{cases} c(1-x^2) & -1.5 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}$$

I'm attempting to understand the intuition of applying the cumulative distribution function from above question question.

I made use of following questions/answers/references :

to find the distribution function

Find C and the distribution function

https://en.wikipedia.org/wiki/Probability_density_function

The cumulative distribution function will be ? :

$$F(x)=\int_0^x{c(1-x^2)}dx$$

From the cumulative distribution function how to find find $P(-.5 \lt X \le .75)$ ?

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To avoid confusion, use $F(x)=\int_0^x c(2-s)^3\mathsf d s$

Now, $\int c (2-s)^3\mathsf ds=\dfrac{-c (2-s)^4}{4}$, so

$$F(x)=\dfrac{-c}{4}{\Big[(2-s)^4\Big]}_{s=0}^{s=x}\mathbf 1_{x\in(0;\underset{?}{\tiny\boxed{3}})}+\mathbf 1_{x\in [\underset{?}{\tiny\boxed{3}};\infty)}$$


Though, as DrHab notes, the given function is negative over the latter part of its domain, so cannot be a probability density function. That 3 is likely a typographical error; perhaps it is mean to be 2?.