TRAIN_CDF_ch6 and bivaraiate class ex

(See md for full stuff ltr)

The Problem

$$\begin{array}{c}f(y)=\{\begin{array}{ll}\frac{c}{500}(25-y^2)& \text{if }-5<y<5\\ 0& \text{elsewhere}\end{array}\end{array}$$

Solution

Then its as simple as finding the $\int$ for this. We know that over $[-5,5]$, the AUC must be = 1, more formally we have the top part of f(y) equating to 1

Firstly we want to find the value of $\frac{c}{500}$ (which is a bit weird since we want to find C = c in our integral)

$$\begin{array}{c}a=\frac{c}{500}\\ a{\int }_{-5}^{5}25-y^{2}dy=1\\ a{\left[25y-\frac{y^3}{3}\right]}_{-5}^5=1\\ a\left(\frac{500}{3}+C\right)=1\\ \frac{c}{500}\left(\frac{500}{3}\right)=1\\ \frac{c}{3}=1,c=3\end{array}$$

So we can now solve the actual integral up to any given point $i$ as such:

$$\begin{array}{c}\frac{3}{500}{\int }_{-5}^{i}25-y^{2}dy=1\\ \frac{3}{500}{\left[25y-\frac{y^3}{3}\right]}_{-5}^i\\ \frac{3}{500}\left(\left(25i-\frac{i^3}{3}\right)-\left(25(-5)-\frac{(-5{)}^3}{3}\right)\right)\\ \frac{3}{500}\left(25i-\frac{i^3}{3}-\left(-125-\frac{-125}{3}\right)\right)\\ \boxed{\frac{3}{500}\left(25i-\frac{i^3}{3}+\frac{250}{3}\right)}\\ \text{we realise that simply simplfying the 3/500 into the rest will match our orignal "answer"}\end{array}$$