Video Transcript
Let 𝑎, 𝑏, and 𝑐 be real numbers
selected randomly from the interval zero, one. What is the probability that the
equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero has at least one real solution
for 𝑥? Rounding the value to four decimal
places.
At first glance, it’s difficult to
see what this problem has to do with integration. But consider the tuple of random
numbers 𝑎𝑏𝑐, where a, 𝑏, and 𝑐 are all between zero and one. We can think of this tuple as a
coordinate in 3D space. Consider a Cartesian coordinate
system with the three axes labeled 𝑎, 𝑏, and 𝑐. And consider the cube of 3D space
that encloses the interval of 𝑎, 𝑏, and 𝑐 between zero and one. This cube contains all possible
combinations of 𝑎, 𝑏, and 𝑐 that we may choose from randomly for the
equation. So, for example, the point 0.1,
0.2, 0.3 lies within this cube and is a valid choice for 𝑎, 𝑏, and 𝑐.
Now, consider the equation
itself. It is a general quadratic equation
in 𝑥. So, like any quadratic equation,
its roots may be given by the quadratic formula. So 𝑥 is equal to negative 𝑏 plus
or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. The key part of this formula that
determines the nature of the solutions is the expression 𝑏 squared minus four 𝑎𝑐,
also known as the discriminant. The value of this expression will
determine whether the equation has two complex roots, one real root, or two real
roots. If the discriminant is less than
zero, it will have two complex roots. If it is equal to zero, it will
have one real root. And if it is greater than zero, it
will have two real roots.
We are asked to find the
probability that the equation has at least one real solution for 𝑥. So we are interested in the two
cases where the discriminant is equal to zero or greater than zero. This of course simplifies to the
inequality 𝑏 squared minus four 𝑎𝑐 is greater than or equal to zero.
Going back to our unit cube in the
3D coordinate system, there will be points within this cube 𝑎𝑏𝑐 which satisfy
this inequality and points that do not. Now, let’s consider the equation 𝑏
squared equals four 𝑎𝑐. This equation defines a surface in
3D space. If we orient the system such that
𝑏 is the vertical axis, as we have done, then anywhere above this surface 𝑏
squared will be greater than four 𝑎𝑐 and anywhere below it 𝑏 squared will be less
than four 𝑎𝑐. So any coordinate 𝑎𝑏𝑐 above this
surface defined by 𝑏 squared equals four 𝑎𝑐 but still within this cube defined by
the interval zero, one satisfies both conditions that 𝑎, 𝑏, and 𝑐 lie between
zero and one and 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero has at least one real
solution of 𝑥. So we need to find the probability
that any random point 𝑎𝑏𝑐 that we choose in the open interval zero, one also lies
above this surface.
Let’s call the entire volume
enclosed by the cube 𝑣 and the volume enclosed by both the cube and above the
surface 𝑏 squared equals four 𝑎𝑐 𝑣 prime. So the probability that 𝑎𝑏𝑐 lies
in 𝑣 prime given that it lies in 𝑣 is just equal to the ratio of the two volumes,
𝑣 prime over 𝑣. This is a unit cube, so the entire
volume 𝑣 is just equal to one. So we need only find 𝑣 prime and
we have our answer. But to find 𝑣 prime, we will need
to use triple integration to find the volume enclosed above the surface and within
the cube.
Let’s clear some space before we
continue. So to find 𝑃, we will need to find
the triple integral over 𝑣 prime of the volume element d𝑣. We can go about this triple
integral in any order that we choose. But we will have to be extremely
careful with the limits. Let’s consider a horizontal slice
of 𝑣 prime with constant 𝑏. If we look straight down the
𝑏-axis at this horizontal slice, then the boundary of the surface 𝑏 squared equals
four 𝑎𝑐 looks something like this in the 𝑎𝑐-plane. Since this is a horizontal slice,
𝑏 is constant and therefore 𝑏 squared is constant.
We can rearrange the equation of
this curve to give 𝑎𝑐 equals 𝑏 squared over four, which is still a constant. So we have 𝑎𝑐 equal to a
constant. So this curve is a hyperbola in the
𝑎𝑐-plane. As we move below or to the left of
this curve, 𝑎𝑐 will decrease. And as we move above or to the
right of this curve, 𝑎𝑐 will increase. So below and to the left of this
curve, 𝑏 squared is greater than four 𝑎𝑐. And above and to the right of this
curve, 𝑏 squared is less than four 𝑎𝑐.
The region we’re interested in is
where 𝑏 squared is greater than or equal to four 𝑎𝑐, so below and to the left of
this curve. We therefore need to find the area
of this shaded region here below and to the left of this curve. Once we’ve done that, we’ll have
the vertical cross-sectional area of this horizontal slice of 𝑣 prime, which we
will then be able to integrate vertically to find the volume 𝑣 prime.
We need to take extra care with the
limits, however, since neither 𝑎 nor 𝑐 can exceed one. So we need to find the restricted
area beneath 𝑎 equals one and 𝑐 equals one. Let’s call the area of this
horizontal slice 𝛼. We can go back to our equation for
the probability 𝑃 and reexpress this triple integral as a single integral between
𝑏 equals zero, the bottom of the cube, and 𝑏 equals one, the top of the cube, of
𝛼, the cross-sectional area of the horizontal slice, with respect to 𝑏.
So first, we need to find an
expression for 𝛼 in terms of 𝑏. We can start by splitting this area
𝛼 into little strips with area d𝛼. These strips have a width of 𝑐 and
a height of d𝑎. We can now integrate these strips
between the limits of 𝑎 to find the area 𝛼. But we need to take extra care with
the limits of 𝑎, particularly when we get to this point. At this point, 𝑐 is equal to one
and we cannot exceed this value since this is outside of the interval zero, one. Therefore, there is a hard limit on
the integration here. At this point, from our equation of
the curve, 𝑎 is equal to 𝑏 squared over four. So once we’ve integrated down to
this point, we need to add on the area of this rectangle here, which is just 𝑏
squared over four times one, so just 𝑏 squared over four.
So let’s summarize what we have so
far. For the vertical cross-sectional
area of this horizontal slice 𝛼, we have the integral between 𝑎 equals 𝑏 squared
over four and 𝑎 equals one of 𝑐 d𝑎 plus the area of this rectangle at the end
here, 𝑏 squared over four.
We will next need to express 𝑐 in
terms of 𝑎 and 𝑏 in order to continue. This is straightforward because 𝑐
is given by the equation of the curve. So 𝑐 is equal to 𝑏 squared over
four 𝑎. So substituting this in, we have
the integral of 𝑏 squared over four 𝑎 d𝑎. In the context of this integral, 𝑏
squared over four is a constant. So we can take it outside the
integration. So this gives us 𝑏 squared over
four times the integral between 𝑎 equals 𝑏 squared over four and 𝑎 equals one of
d𝑎 over 𝑎 plus 𝑏 squared over four.
So, integrating, we have 𝑏 squared
over four times the natural log of 𝑎 evaluated between 𝑏 squared over four and one
plus 𝑏 squared over four. Evaluating, the natural log of one
is just equal to zero. So this simplifies to 𝑏 squared
over four minus 𝑏 squared over four times the natural log of 𝑏 squared over
four. We will not factorize this just yet
as the next integral will be much simpler if we don’t. We can now insert this expression
for 𝛼 back into our main integral for the probability. Doing this and clearing some space,
we get 𝑃 is equal to the integral between 𝑏 equals zero and 𝑏 equals one of 𝑏
squared over four minus 𝑏 squared over four times the natural log of 𝑏 squared
over four with respect to 𝑏.
This first term in the integrand is
just a polynomial, so this is easy. This second term, however, will
require a little more thought. To better keep track of things,
let’s split this integral into two separate integrals between zero and one of 𝐼 one
with respect to 𝑏 minus zero and one of 𝐼 two with respect to 𝑏, where 𝐼 one is
𝑏 squared over four and 𝐼 two is 𝑏 squared over four times the natural log of 𝑏
squared over four. Let’s start with the second
integral. We’re integrating some function of
𝑏, namely, 𝑏 squared over four, multiplied by the natural log of that same
function.
Consider the general form of this
scenario, the integral of some function 𝑓 of 𝑏 times the natural log of that same
function. Let’s use integration by parts. Recall that if we have the integral
of some composite function 𝑢𝑣 prime, then this is given by 𝑢𝑣 minus the integral
of 𝑣𝑢 prime. If we let 𝑢 equal the natural log
of 𝑓 of 𝑏 and 𝑣 prime equal 𝑓 of 𝑏, then 𝑢 prime is given by 𝑓 prime of 𝑏
over 𝑓 of 𝑏 and 𝑣 is given by the integral of 𝑓 of 𝑏 with respect to 𝑏. Substituting this into our formula,
we get the integral of 𝑓 of 𝑏 times the natural log of 𝑓 of 𝑏 with respect to 𝑏
is equal to the natural log of 𝑓 of 𝑏 times the integral of 𝑓 of 𝑏 with respect
to 𝑏 minus the integral of the integral of 𝑓 of 𝑏 with respect to 𝑏 times 𝑓
prime of 𝑏 over 𝑓 of 𝑏 all with respect to 𝑏.
In our case, 𝑓 of 𝑏 is equal to
𝑏 squared over four, which means that 𝑓 prime of 𝑏 is equal to 𝑏 over two. And the integral of 𝑓 of 𝑏 with
respect to 𝑏 is 𝑏 cubed over 12. Plugging all of these in, we get
the integral between zero and one of 𝐼 two with respect to 𝑏 is equal to the
natural log of 𝑏 squared over four times 𝑏 cubed over 12 evaluated between zero
and one minus the integral between zero and one of 𝑏 cubed over 12 times 𝑏 over
two all over 𝑏 squared over four with respect to 𝑏.
This second integrand simplifies to
just 𝑏 squared over six. So integrating gives us just 𝑏
cubed over 18 evaluated between zero and one. This second term evaluates easily
to just one over 18. But this first term presents a
problem since we will be evaluating at 𝑏 equals zero. So we will be taking a natural log
of zero, which is undefined. We can, however, get around this by
finding the limit as 𝑏 tends to zero of this term. If we take the reciprocal of this
𝑏 cubed and place it on the denominator, then we can rewrite this limit as the
limit as 𝑏 tends to zero of the natural log of 𝑏 squared over four all over 12𝑏
to the negative three. As 𝑏 tends to zero, the numerator
will tend to negative ∞ and the denominator will tend to ∞. So we have an indeterminate
limit.
Let’s clear some space and move
this down here before continuing. Since we have an indeterminate
limit, we can use L’Hôpital’s rule, which states briefly that if we have an
indeterminate limit of this form, 𝑓 of 𝑥 over 𝑔 of 𝑥 as 𝑥 tends to 𝑐, then
this is equal to the limit as 𝑥 tends to 𝑐 of 𝑓 prime of 𝑥 over 𝑔 prime of
𝑥. So in our case, our 𝑓 of 𝑥, or 𝑓
of 𝑏 in this case, is the natural log of 𝑏 squared over four and our 𝑔 of 𝑏 is
12 times 𝑏 to the negative three. So plugging in our expression to
L’Hôpital’s rule and taking the derivative of the numerator and the denominator, we
get the limit as 𝑏 tends to zero of two over 𝑏 all over negative 36 times 𝑏 to
the negative four. This expression simplifies to
negative two-thirds 𝑏 cubed. And so the limit as 𝑏 tends to
zero of negative two-thirds 𝑏 cubed is just equal to zero.
This means that this term in the
integral evaluated at zero is just equal to zero. So we can now evaluate this whole
integral equal to one twelfth times the natural log of one-quarter minus one
eighteenth. Substituting this back into our
equation for 𝑃, we now have 𝑃 is equal to the integral between zero and one of 𝑏
squared over four with respect to 𝑏 minus one twelfth times the natural log of
one-quarter minus one eighteenth. Integrating this first term, we get
𝑏 cubed over 12 evaluated between zero and one. Evaluating and simplifying gives us
our probability equal to five over 36 minus one twelfth times the natural log of
one-quarter. And in decimal form, this is 0.2544
to four decimal places.