### Video Transcript

Let π be a continuous random
variable with probability density function π of π₯ is equal to ππ₯ if one is less
than or equal to π₯ is less than or equal to five and zero otherwise. Determine the value of π.

In this example, weβre given that
π of π₯ is a probability density function of a continuous random variable. And weβre asked to find the
constant π. To do this, letβs recall the
properties of a probability density function. π of π₯ is a probability density
function if π of π₯ is greater than or equal to zero for all values of π₯ and the
total area under the graph of π¦ is equal to π of π₯ is equal to one. So letβs begin by examining the
first condition; thatβs the positivity condition.

We know that π of π₯ is equal to
zero outside of the interval one to five. And this means that our first
condition is satisfied for π₯ not in the interval one to five. For π₯ inside the interval one to
five, we know that π of π₯ is equal to ππ₯. We know that π₯ is greater than
zero in this interval. And so π also must be greater than
or equal to zero, since π is a probability density function. Our first condition then requires
that π is greater than or equal to zero. Now letβs look at our second
condition, which tells us that the total area under the graph of π¦ is π of π₯ is
one. In order to satisfy this condition,
we see that π cannot be equal to zero, since if π is equal to zero, π of π₯ is
equal to zero for all π₯. This in turn means that the area
would be equal to zero.

Hence, to satisfy the second
condition that the total area is equal to one, we cannot have π equal to zero. This means that π must be greater
than zero. And since π of π₯ is equal to ππ₯
for π₯ between one and five, the graph of π of π₯ over the interval one to five
must be a straight line with a positive slope. Now, from this graph, we see that
the area under the graph is a trapezoid. And we recall that the area of a
trapezoid is one over two multiplied by the sum of the lengths of the base and the
top multiplied by the height. And so to find the area of this
trapezoid, thatβs the area under the curve, we must find the lengths of the base,
the top, and the height.

To do this, we begin by finding the
coordinates of the vertices of the trapezoid. And to do this, we substitute π₯ is
equal to one and π₯ is equal to five into our function π of π₯. π of one is equal to π multiplied
by one, which is equal to π. And π of five is equal to π
multiplied by five, which is five π. And this gives us the coordinates
of our vertices, which are one, π and five, five π. And so the top of our trapezoid has
a length π and the base has a length five π.

The height of our trapezoid is the
length that lies along the π₯-axis. That is five minus one, which is
four. And so we have the top π, the base
five π, and the height four, which we can now substitute into the formula for the
area of the trapezoid. And so we have the area is one over
two multiplied by five π plus π multiplied by four. That is two multiplied by six π,
which is 12π.

Now to satisfy our second
condition, 12π must be equal to one because thatβs the area. Dividing both sides by 12, we can
solve for π. And so the value of π is equal to
one over 12.