Welcome to lesson on
the area between two curves.
For review, if F of X is non-negative,
then the def integral of F of X
integrated with respect to X from A to B
would give us the area bounded by the function
and the X axis.
So looking at our graph here,
if we integrated this blue function from zero to six,
notice how the function is non negative on this interval,
and therefore the def integral
would give us the area of the shaded region.
So to find the area bounded by two curves
we just extend this idea.
If F of X and G of X are continuous
on the close interval from A to B,
and G of F is less than or equal to F of X
for all X in the interval,
then the area of the region bounded
by the two graphs of F and G,
and the vertical lines X equals A and X equals B
is given by the def integral of F of X
minus G of X integrated with respect to X from A to B.
So one of the most important things to recognise here,
is when we set this up, F of X must be the top function
and G of X must be the bottom function.
So looking at our graph over here on the right,
if we wanted to find the area bounded by these two functions
or the area of this blue shaded region,
if we began by considering the points of intersection.
Which are here and here.
Notice how if we found the area under the quadratic function
over this interval and then subtract the area
under the linear function over the same interval,
notice how it would leave us with the area
of this blue shaded region.
The area bounded by the two curves.
So to find the area bounded by two intersecting curves,
we will need to calculate the points of intersection
by setting the two functions equal to each other.
The upper and lower limits of integration
will be determined by the points of intersection.
Looking at the graph on your calculator,
or some other graphing software is often very helpful,
and sometimes you may have more than
two points of intersection.
Lets take a look at an example.
Here's the graphic considered on the previous slide,
where we want to find the area bounded by
Y equals 2x which is graphed here in red,
and Y equals 6x minus X squared graphed here in blue.
So again we're trying to find the area
of this blue shaded region.
So the first step I should determine
the points of intersection,
and because we have the graph,
we can tell this point of intersection would be the origin
0,0 and this point of intersection would be the point 4,8.
So notice how we will be integrating from zero to four,
but let's also find these points
of intersection algebraically.
To do this, we would set the two functions
equal to each other.
So we'd have 2x equals 6x minus X squared.
Notice how we have a quadratic equation
so we'll set it equal to zero and see if we can factor.
So we'll subtract 2x on both sides.
6X minus 2x is 4x,
and notice how this does factor.
We have a common factor of X.
So we have X times the quantity of four minus X.
So this would equal zero.
When X equals zero or when four minus X equals zero.
So here if we add X on both sides,
we have four equals X or X equals four,
and here we already know X equals zero.
So this does verify our limit of integration
will be from zero to four.
So the area can be equal to the integral of,
here we'll have the top function minus the bottom function.
So notice how we have the quadratic function
minus a linear function.
So we'd have six X minus X squared
minus our linear function is Y equals two X.
Integrated with respect to X from zero to four.
Let's go and simplify the integrand.
Here we have six X minus two X, that's four X.
Now we'll find the anti derivative.
So we'd have four times, that's X to the first
so we'd have X to the second divided by two
minus X to the third divided by three.
Let's continue on the next slide.
Simplifying here we have two X squared.
Let's write this as minus one third X to the third.
So when X is four we have two times four squared
minus one third times four cubed.
In notice when X is zero, both terms are zero.
So here we have four squared is 16, times two, that's 32.
Minus, here we have one third times
four to the third, 64 thirds.
Obtaining a common denominator.
Which would be three.
We have 96 thirds minus 64 thirds or 32 thirds
as the area bounded by these two curves.
Which is this blue shaded region.
Let's take a look at another example.
We want to find the area bounded by Y equals X
and Y equals five X minus X cubed.
So we're trying to find the area of these
two shaded regions.
Notice how these two functions do have three
points of intersection,
and also on the left, the linear function is on top
and the quadratic is on the bottom,
and on the right, the quadratic function is on top
and the linear function is on the bottom.
Which means to find these two areas,
we'll have to set up two separate integrals.
However, in this case because of the symmetry
of the two areas, we could find one of the areas
and then just double the area,
but just in case we didn't have symmetry,
I do want to set this up as two separate integrals.
Okay I'm looking at the graph,
we can tell that the points of intersection
would be -2,-2 here.
The origin 0,0 here, and the .2,2 here,
but again I do want to verify the X coordinates
of these points of intersection algebraically first
by setting these two functions equal to each other.
So to do that algebraically we would have,
when is X equal to five X minus X cubed.
If we subtract X on both sides we'd have zero equals four X
minus X cubed, and this does factor.
Common factor of X again, getting us with X times
the quantity four minus X squared.
Here we have the difference of squares.
So we have X and then we have two binomial factors.
Four's a perfect square, we'd have two times two.
X squared a perfect square, we'd have X and X,
and we'd have a sum and a difference.
So this part equals zero when X equals zero,
X equals negative two or X equals positive two.
Notice how these are the X coordinates
of the point of intersections.
Which will give us the limits of integration.
So again, if I needed the area separately
and not using the symmetry,
we would have the area of the bounded region
is equal to the integral of, on the left we have
the linear function on top, the quadratic on the bottom.
So we would have X minus the quantity five X minus X cubed
integrated with respect to X from negative two to zero,
and then plus the integral of, on the right
the quadratic's on top, so we would have
the quantity five X minus X cubed minus X
integrated with respect to X from zero to two,
and again, because of the symmetry we could find
one of the areas and then just double that value,
but let's go ahead and solve it this way.
So next we'll simplify the integrams.
So here we have X minus five X.
That would be negative four X.
Then we have minus negative X cubed,
that's plus X cubed
plus the integral from zero to two.
This is going to be the opposite.
We have five X minus X, that's four X,
and then minus X cubed.
Let's continue on the next slide.
So now we'll find the anti derivatives.
So we'd have negative four times X squared divided by two
plus X to the fourth over four,
and then plus, here we have positive four times
X squared over two minus X to the fourth over four.
Let's go and simplify these.
So here we have negative two X squared.
Let's write this as plus one fourth X to the fourth,
and here we'll have positive two X squared
minus one fourth X to the fourth.
So here we'll substitute zero first.
Notice when X is zero both terms are zero,
and then when X is negative two
we'd have negative two times negative two squared
plus one fourth times negative two to the fourth,
and then here when X is two we have two times two squared.
Then we have minus one fourth two to the fourth
and when X is zero both terms are zero.
So finally simplifying we'd have negative,
and then here we have positive four times negative two,
that's negative eight,
and here we'd have negative two to the fourth.
That's 16 times one fourth, that's four,
and we have plus.
Here we have eight minus, again, 16 times one fourth is four
so we have eight minus four.
So here we have the opposite of negative four,
and then plus four.
Which is four plus four or eight.
So this area is four square units.
This area is four square units.
So the combined bounded area is eight square units.
I hope you found this helpful.