Finding the height and probability of a probability density function 

*A continuous random variable X has a probability density function f(x) represented on the diagram below (0’A’B’C’). 

A) FIND THE VALUE OF H
We know that the total area must be equal to 1. We can divide the probability density function into two different sections that can be easily calculated (rectangle and a triangle). Given this, we know that we know that the total area is the sum of the area of the rectangle and the area of the triangle. 
We have both the height, and the width needed to calculate the area of the rectangle
Area rectangle = 1/4 · 3 = 3/4
We don’t have the height for the triangle, so we solve with reference to total area
1 =  3/4 + Area triangle
Area triangle = 1/4
Using the area of the triangle, we solve for height h
1/4 = ((h-1/4)2)/2 = h - 1/4
h = 1/4 + 1/4 = 1/2
B) COMPUTE P (0 < X ≤ 1) 
The probability of P (0 < X ≤ 1) is equivalent to the area under the curve from 0 to 1. This area is a rectangle so we use the formula A = hb to find the area. 
A= (0.25)(1) = 1/4
C)COMPUTE P (1 < X < 2) 
I'm really not sure how to answer this one, any help would be appreciated. 
 A: We compute $\Pr(1 < X < 2)$ by computing the area under the curve from $x=1$ to $x=2$. This can be calculated as the sum of the areas of a rectangle and a triangle. 
The area of the rectangular region is $0.25 \times 1=0.25$.
For computing the area of the triangle, we use the properties of similar triangles. In particular, the height of the triangle is $\frac{h}{0.25}=\frac{1}{2} \implies h=1/8$. Consequently, the area of the triangle is $\frac{h}{2}=\frac{1}{16}$.
Therefore,
$$\Pr(1 < X < 2) = \frac{1}{4}+\frac{1}{16}=\frac{5}{16}.$$
A: $\mathsf P(1<X<2) = \frac 12(\lvert AA'\rvert+\lvert BB'\rvert)\cdot \lvert AB\rvert$ from the area of a double-right-angled trapezium.
You should find that the height $\lvert BB'\rvert$ is $\frac{(\lvert AA'\rvert+\lvert CC'\rvert)\cdot\lvert AB\rvert}{\lvert AC\rvert}$ by reason of similar triangles.
A: The middle line of the trapezium $AA'C'C$ is
$$BB'=\frac{AA'+CC'}{2}=\frac{\frac14+\frac12}{2}=\frac38.$$
The area of the trapezium $AA'B'B$ is
$$P(1<X<2)=\frac{AA'+BB'}{2}\cdot AB=\frac{\frac14+\frac38}{2}\cdot 1=\frac{5}{16}.$$
