Question 1082961
Given piecewise function
<img src="https://i.imgur.com/bUg2uou.png">
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To find the CDF function F(x), we integrate each piece of the piecewise function f(x).
For the first piece:
*[Tex \LARGE \int_{0}^{k} x dx = \left.\frac{x^2}{2}+C\right|_{0}^{k}]
*[Tex \LARGE \int_{0}^{k} x dx = \left(\frac{k^2}{2}+C\right)-\left(\frac{0^2}{2}+C\right)]
*[Tex \LARGE \int_{0}^{k} x dx = \frac{k^2}{2}]
where 0 < k <= 1
So the first piece of F(x) is {{{(x^2)/2}}} where 0 < x <= 1.
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Repeat for the other piece
*[Tex \LARGE \int_{1}^{m} (2-x) dx = \left.2x - \frac{x^2}{2} + C\right|_{1}^{m}]
*[Tex \LARGE \int_{1}^{m} (2-x) dx = \left(2m - \frac{m^2}{2} + C\right)-\left(2(1) - \frac{1^2}{2} + C\right)]
*[Tex \LARGE \int_{1}^{m} (2-x) dx = -\frac{m^2}{2} + 2m-\frac{3}{2}]
where 1 < m <= 2
Since the area is cumulative, we need to include the green shaded triangle (shown below) that has area 1*1/2 = 1/2
<img src = "https://i.imgur.com/mTSUXGn.png">
This green triangle area is represented by F(x) when x = 1.
It comes from the first part of the piecewise function F(x)
Add this on to the integral result to get
{{{(previous_result)+1/2 = ((-m^2)/2 + 2m - 3/2) + 1/2 = (-m^2)/2 + 2m - 1}}}
making the second piece of F(x) to be {{{(-x^2)/2 + 2x - 1}}} when 1 < x <= 2.
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So overall, the CDF function is
<img src="https://i.imgur.com/w2DoDBp.png">
which is the final answer
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Extra Information:
Graph of the piecwise function f(x) in red, and the piecewise CDF function F(x) in blue
<img src="https://i.imgur.com/lGwmHRU.png">
Image generated by <a href = "https://www.geogebra.org/home?ggbLang=en">GeoGebra</a> (free graphing software).


As you can see, f(x) forms triangle. The base is 2 and the height is 1. The area under all of f(x) is therefore
A = base*height/2
A = 2*1/2
A = 2/2
A = 1
Since the area under the function curve f(x) is 1, this confirms that f(x) is indeed a Probability Distribution Function (PDF).