# Find area integral

Given the line $y = 2x$, I need to find the area of the region below it.

Original question:

• Two numbers $x$ and $y$ chosen randomly from interval $\left[0,1\right]$, find the probability that $2x > y$.
• In order to prove that, I need to find the area of the region under the line $y = 2x$. The answer is $3/4$. How can I get this result ?.
• The support of $(X, Y)$ is a unit square. As you said $2x = y$ is a straight line. And this line intersect with the square and form a triangle. And thus...
– BGM
Jan 5, 2018 at 8:42
• Base by height divided by 2?
– N74
Jan 5, 2018 at 8:42
• The answer is 3/4. How? Jan 5, 2018 at 8:43

First, imagine your input space as being in the first quadrant, where there's a unit square with vertices $(0,0), (1,0), (1,1), (0,1)$. Draw this out. Now draw the line $y=2x$. If you pick any point $(x,y)$ where $x \in [0,1], y \in [0,1]$, it will fall somewhere in this unit square. You need to figure out what part in this square satisfies $y < 2x$. I find it easier to work with the probability that $y \geq 2x$ and then you can easily find the other probability. So if we're trying to find the probability that $y \geq 2x$, you shade the upper part of the triangle bounded by the line $y = 2x$ and the top of the unit square ($y = 1$). Color this triangle in.
You can now compute where the line $y=2x$ intersects the line $y=1$. From here, you now have the width and height of this colored-in triangle. Compute the area of that, and divide by the area of the unit square. This gives you the probability, which comes out to be $.25$. This is the probability, remember, that $y \geq 2x$. To find the complement of this probability, that $y < 2x$, you just do $1 - .25 = .75$.
In fact you need to use not only the fact that $x\in [0,1]$ but also the fact that $y\in [0,1]$.
Calculate the total area of the region $x,y\in [0,1]$: it is $1\times 1=1$. Now we have to calculate the total area below the graph of $y=2x$ in the region $x,y\in [0,1]$. This is the tricky part since it is in fact not a triangle but a trapezoid. To see this, calculate $y(x=1)=2\times1=2$, which is bigger than $1$, the maximum of $y$ in our region. We therefore use height $1$ for our trapezoid. One base length is just the interval $x\in [0,1]$ whose length is $1$. The other base is the interval between the boudary of our region $(1,1)$ and the intersection of the lines $y=1, y=2x$. That intersection is $(\frac12,1)$, and therefore the length of this base is $1-\frac12=\frac12$. The area of the trapezoid is therefore $A=\frac12((1+\frac12)\times1)=\frac34$, and the probability is the area of the trapezoid divided by the area of the region $P=\frac{\frac34}{1}=\frac34$, as requested.