7.1a Area of a Region between
•Find the area of a region between two
curves using integration
•Find the area of a region between
intersecting curves using integration
•Describe integration as an accumulation
With just small modifications you can extend the idea of
definite integrals (area under a curve and above (or below) the
x-axis) to area between two curves.
Consider f(x) and g(x) that are continuous on an interval [a, b]
where the graph of g lies below that of graph of f
This representative rectangle
is a HUGE idea. It is the
basis for your entire setup, so
hopefully it makes sense to
you. Right now it seems
trivial, but if you get into the
habit of sketching in a
volumes of revolution (7.2
and 7.3) will make sense.
Area = height ∙ width
We’re just adding up areas
As you can see from
the graphs, it
doesn’t really matter
where the two
graphs are on the
plane, as long as f(x)
is ABOVE g(x) for
the interval [a, b]
Representative rectangles are used throughout this chapter. A vertical rectangle
(of width Δx) implies integration with respect to x. A horizontal rectangle (of width Δy)
implies integration with respect to y.
Ex 1 p447 Finding the Area of a Region Between Two Curves
Find the area of the region bounded by the graphs of
1, , 0, and 1y x y x x x= + = − = =
[( 1) ( )]x x dx+ − −∫ dx: adding up rectangles from
left to right of width Δx
Left-most value of x
Right-most value of x
= + +
Top function Bottom function
In the last example, values for a and b have to be given explicitly. A more common
problem occurs if the graphs intersect and you must calculate the values for a and b .
Ex 2 p. 448 A Region Lying Between Two Intersecting Graphs.
( ) 1, g(x) 1f x x x= − = −
Solution: Graph both and find intersection
points. The x-values will be the interval we
are putting for lower and upper limits of
integration. Draw in representative rectangle.
Then set up integral:
[ 1 ( 1)]x x dx− − −∫
= − − +
Ex 3 p. 448 A Region Lying Between Two Intersecting Graphs
Since the sine and cosine curves intersect infinitely many times, you are to find
only one such region.
Find the region bound by f(x) = sin x and g(x) = cos x
I decided to look at region from x = -3π/4 and x = π/4
What function is on top in this interval?
cos sinx x dx
−∫ ] 4
sin ( cos )x x
= − −
If you are solving by calculator, make sure you realize that you are looking
for f(x) – g(x)! Not just area under the last curve you graphed, which might
be just g(x) or just f(x).
If two curves intersect at more than two points, then you must split up into more than
one interval based on the x-values of ALL the points of intersection. Make sure you
realize which graph is on top in a particular interval.
Ex 4 p. 449 Curves that intersect at more than two points.
Find the area of the region between the graphs of
f(x) = 3x3
– 10x and g(x) = -x2
f(x) = g(x) to find intersections: set = 0, then solve.
x = 0, 2. Which function is on top for [-2, 0]? Which is
on top for [0, 2]? Be careful with algebra here!
3 2 2 2 3 2
[(3 10 ) ( 2 )] [( 2 ) (3 10 )]x x x x x dx x x x x x dx
− − − − + + − + − − −∫ ∫
(3 12 ) ( 3 12 )x x dx x x dx
− + − +∫ ∫
Since these are complicated enough, I will let you find the
definite integral by calculator. However, you must show what
integrals in simplest form you are entering with correct limits.
12 12 24= + =
It is a coincidence that both regions are the same area!
You would have gotten the wrong result if you had merely integrated
f(x)-g(x) from [-2, 2]
Ex 5 p.450 Horizontal Representative Rectangles
Find the area of the region bounded by the graphs of x = 3 – y2
and x = y + 1.
(Note: these could also have been written f(y) = 3 – y2
and g(y) = y + 1)
You have a couple of choices when trying to visualize them:
Graph by hand, or graph the inverse functions by calculator,
turn them off, and then go to DRAW menu, choose DrawInv
and first select VARS FUNCTION Y1 and then repeat for Y2
Notice that the representative rectangle is horizontal, so the widths will by ∆y and
the heights will be the difference between the function further to the right and the
one further to the left! You will also need to find points of intersection: Set x=x or
in other words, 3 – y2
= y + 1 and solve for y. This leads to (y+2)(y-1)=0, and
y = -2, 1
[(3 ) ( 1)]y y dy
− − +∫
( 2)y y dy
= − − +∫
= − + =
Notice that we just worked with one integral here!
With this particular graph, we can choose to integrate
with respect to y, like we just did, or with respect to x,
as shown here or on p.450. Vertical rectangles with
respect to x is MUCH more of a pain!
With some setups, it is impossible to even do this. So
get your mind used to seeing the problems as a bunch
of vertical rectangles, or as a bunch of horizontal
rectangles, whichever gives you easier/possible
For example, how would
you even do a problem like
finding the area between
f(y) = y3
-3y and g(y) = y
from y = 0 to y = 2 as