Question
Show that the function is a probability density function on the specified interval.$$f(x)=\frac{2}{9}\left(3 x-x^{2}\right) ;(0 \leq x \leq 3)$$
Step 1
This is done to find the total area under the curve of the function in the interval [0,3]. The integral is given by: $$\int_{0}^{3} f(x) dx = \int_{0}^{3} \frac{2}{9}\left(3 x-x^{2}\right) dx$$ Show more…
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