It’s Professor Dave, let’s find the area between curves.
By now, we have a solid conceptual understanding of integration, and are able to integrate
a wide variety of functions.
So it’s time to start applying integration to different kinds of problems in order to
solve them.
Let’s dive into these applications now.
To set ourselves up for the first type of problem we are going to examine, take a look
at this figure.
It’s a square with a circular hole in it.
If we wanted to find the area of this figure, how would we do it?
Well if we knew all of these parameters, it would be quite simple.
We could just find the area of the circle, and subtract it from the area of the square,
because the circular area is being taken away from the square’s area.
By that analogy, take a look at this.
Here are two continuous functions, f of x and g of x, operating over the same interval.
How would we find the area in between these two curves?
Well if we learned anything from our little square donut, we can use exactly the same
approach.
We can find the area in between f and the x-axis by integrating f of x.
We can find the area between g and the x-axis by integrating g of x.
So if we only want to know about this section between these two curves, then we should be
able to subtract the integral of g of x from the integral of f of x.
That would be subtracting the smaller area from the total area, just like we subtracted
the circle from the square.
This is a powerful concept, because prior to this, we had been exclusively calculating
the area between a curve and the x-axis, but simply by subtracting two integrals, we can
now find the area between any two functions, anywhere on the coordinate plane.
Let’s try an example.
Here are two functions, the first being the parabola x squared plus one, and the other
being the line y equals x.
Let’s find the area in between these curves over the interval from zero to one.
Rather intuitively, that will be the integral of x squared plus one, from zero to one, minus
the integral of x, from zero to one.
Again, that’s like taking all of this area, and subtracting from it this smaller area,
which leaves us with the area in between the curves.
Now we know that the difference of two integrals over the same interval is the same as the
integral of the difference of the two functions.
So really, we can even just turn this into one integral.
Then we just end up with the integral of x squared plus one minus x, or rearranged, x
squared minus x plus one.
This may seem incredible, as it’s now an entirely new function, but integrating this
will indeed give us our answer.
So let’s get the antiderivative, which will be x cubed over three, minus x squared over
two, plus x.
We evaluate that at one, and we get one third minus one half plus one.
Changing everything to the common denominator of six, and then combining everything into
one fraction, two minus three plus six is five, so that will be five sixths.
And since evaluating at the lower limit gives us zero, this will be our answer.
So as you can see, this application does not require that we learn anything significantly
new, as we are just integrating functions as we have been doing.
However, we will sometimes have to use a little bit of critical thinking.
What if we are simply given two functions, y equals x squared, and and y equals two x
minus x squared, and we want to find the area of the region enclosed by the two functions.
First things first, let’s graph them so that we can see what we are dealing with.
We know what x squared looks like, and this other one is also a parabola.
We could either use a graphing calculator, or just convert this into vertex form by completing
the square.
If you need a refresher on how to do this, visit my tutorial on this topic earlier in
the series, otherwise, a little bit of rearranging would give us this, so it looks like x squared
but with its center at one, one, and opening up downwards.
So in looking for the area of the region enclosed by these functions, we are certainly implying
this section right here.
So we have the two functions that we need to integrate, but what are the limits of integration?
Well those will be the x values at the points where the functions intersect.
So what are those values?
We can probably tell from this graph, but just to make sure, let’s set the functions
equal to each other and solve, since these points must be solutions for both functions.
A little bit of algebra will give us the origin, and (one, one) as the intersections, just
as we might expect.
So let’s set up our integral, making sure to write two x minus x squared first, and
then subtract x squared from that, as we are subtracting the function that makes up the
lower edge from the function that makes up the upper edge.
This will give us two x minus two x squared.
Integrating this, we therefore get two x squared over two, or simply x squared, minus two x
cubed over three.
Evaluating this expression for x equals one will give us one minus two thirds, which is
one third, and evaluating at zero gives us zero, so one third is our final answer.
So this is the basic idea behind this technique.
The functions can get a little trickier, and we may need to use the substitution method
or some other technique to evaluate the integral, but that won’t change our approach at all,
we are still just subtracting two functions and integrating.
One other thing we may see is a situation where we have functions in terms of y instead,
and in such a situation, we will just be doing everything in terms of the y axis, meaning
that we are gauging how far functions are from the y axis instead of the x axis.
This also means that we will integrate with respect to y rather than x, so y terms are
in the function, and the integrand ends with dy.
But other than this small difference, the approach is absolutely identical.
To get a little more practice, let’s check comprehension.