Welcome to a video on the properties of definite integrals and the area bounded by
two graphs. Let's take a look at some properties first. If we have a definite integral where the
lower limit and the upper limit of integration are the same, that will always equal zero. Hopefully,
that kind of makes sense because the width of the interval we’re considering would be zero.
The second property: if we have a definite integral with the limits of integration from
A to B, changing the direction of integration, meaning we integrate from B to A, would change
the sign of the definite integral. If we have the definite integral of K times a function,
we can factor out the K, and the result would be K times the value of the definite integral.
A couple of other quick properties: if we have a definite integral involving the sum
and difference of two functions, we can break it up into individual definite integrals. Lastly,
if we have a definite integral on the interval from A to B, as we have in this graph here,
it would be okay for us to break this into two different integrals if we integrate from A to
C and then C to B. Hopefully, we can see geometrically that this would be allowed.
Let’s go ahead and focus on the area of a region between two curves or two graphs. Here’s the idea:
again, if f and g are continuous on the closed interval, and g is less than or equal to f of x
for all x in the interval, then the area of the region bounded by the graphs and
the vertical lines x equals A and x equals B is given by this definite integral. The key
to remember here is we always have to have the top function minus the bottom function.
To find the area bounded by two intersecting graphs, we need to calculate the points of
intersection by looking at the graph or by setting the two functions equal to each other.
The upper and lower limits of integration will be determined by the points of intersection.
Let’s give it a try. We want to find the area bounded by these two functions. Step
one is to find the points of intersection, so we're going to set the functions equal
to each other and solve. If we move these two terms to the other side, we’d have a positive
x squared plus x, and then if we subtract three on both sides, we’d have minus two. This does factor,
so we have two points of intersection: we have x equals negative two and x equals one.
These two values will help us find our limits of integration,
but we still have to know which function is on top and which one is on the bottom. So
let's go ahead and take a look at a graph. We can see that f of x is the quadratic equation,
so it's the parabola, and of course, g of x is the line. Here’s our value of x
equals negative two that we found algebraically, and here’s the value of x equals one.
So the area bounded by these two functions will be the definite integral of f of x
minus g of x since f of x is on top on the interval from negative two to
positive one. Let’s go ahead and simplify our integrand. We’d have
x squared minus x and then three minus one or plus two. We will integrate with respect to x. So,
using the power rule, we’d have x cubed over three minus x squared over two plus two x.
We'll evaluate this at one and negative two. If we substitute in positive one, it would
look like this. If we substitute in negative two, it would look like this, and this simplifies to
four point five. So the area bounded by these two functions would be four point five square units.
Let’s go ahead and verify this definite integral on the graphing calculator. If
we hit math and then select option nine, we can type in our original
integrand x squared minus x plus two, comma, x, comma, negative two, comma,
one, just to check our work, and that verifies that we did it correctly.
Let’s go ahead and try another one. Now on this problem, I’m going to go ahead and take
a look at the graph to set this up. I don’t feel like solving this algebraically, but of course,
we can. The first thing we notice on this one is that the blue graph is on top at first and
then the red graph is on top second. This tells us we’re going to have to have two
definite integrals because it always has to be the top function minus the bottom function.
So, I went ahead and labeled these functions so it makes it easier for us to set up our definite
integrals. The first definite integral would be on the interval from negative two to zero,
and we can see that f of x is on top. We’ll label this area one, and we’ll label this area two. For
area two, we’d have the definite integral from zero to two of g of x minus f of x.
Let’s go ahead and simplify these. For area one, we’d have a three x cubed.
Now the x squared terms do cancel, and we’d have negative ten x minus two x or negative twelve x,
and the two also cancels. For area two, we’d have a negative three x cubed. Again,
the x squared terms cancel, and so do the twos. Then we’d have positive twelve x.
Now, the main idea here is setting these up,
so I’m going to go ahead and evaluate these on the graphing calculator.
The first definite integral would be equal to twelve, and the value of the second definite
integral would be equal to twelve as well. Therefore, we take the sum of these two
to find the total area, so the total area would be equal to twenty-four square units.
Now, that is kind of interesting that these two areas are the same even though they are
different shapes. Okay, I hope you found this video helpful. Thank you and have a good day!