Area Between Curves:
Z
b
We know that a definite integral
f (x) dx can be used to find the signed area of the
a
region bounded by the function f and the x−axis between a and b. Often we want to find
the absolute area of a region bounded between two or more curves, and we may use the
definite integral to do this, too. When we are being asked to find the area bounded between
two (or more) curves, we want to draw the graph of the curves and find the point(s) of
intersection(s) if necessary to determine the enclosed region and set up the definite integral
accordingly.
f (x)
g(x)
In the picture above, to find the area of the region bounded between f and g, we use the
method of dividing the interval into rectangles of equal length and try to set up a Riemann
sum. Notice that the height of each rectangle is given by f (x) − g(x). Therefore, when we
use a definite integral to evaluate the area between two curves, the integrand will be the
difference of two functions. The following discussion summarizes the situation we have and
the definite integral that we can set up.
If, inside an interval [a, b], Zf (x) ≥ g(x) for all x, then the absolute area bounded between
b
f (x) − g(x) dx
f and g is given by: A =
a
If the functions intercept and form a closed region, to find the area of the region, we set up
a definite integral with the lower and upper limits of integration as the x−coordinates of
the points of intersection.
f (x)
g(x)
In the graph above, f and g intercept
Z at x = −1 and x = 2, so the area of the region
2
f (x) − g(x) dx
enclosed by f and g is given by: A =
−1
If, inside the interval [a, b], sometimes f is greater, and sometimes g is greater, then the
absolute area of the region enclosed by f and g inside [a, b] can be defined by:
Z
b
|f (x) − g(x)| dx
A=
a
In order to actually evaluate the definite integral, we will have to divide the interval [a, b]
into subintervals (ai , ai+1 ) where inside each subinterval,
either f (x) > g(x) or g(x) > f (x)
Z
ai+1
f (x) − g(x) dx
for all x. If f (x) > g(x), we find the area using:
ai
Z
ai+1
g(x) − f (x) dx
If g(x) > f (x), we find the area using:
ai
In the picture above, f (x) > g(x) for all x inside the interval (a, a1 ) and in (a2 , b); but
g(x) > f (x) for all x inside (a1 , a2 ). To find the absolute area of the region bounded (on
left and right) by x = a and x = b, and bounded (top and bottom) between f and g, we
use:
Z
b
|f (x) − g(x)| dx = S1 + S2 + S3 =
A=
a
a1
Z
Z
a2
f (x) − g(x) dx +
a
Z
b
g(x) − f (x) dx +
a1
f (x) − g(x) dx
a2
Example: Find the area of the region bounded by the curves x = 1, x = 2, y = 4x + 1,
y = ex
Ans: For 1 ≤ x ≤ 2, 4x + 1 is always greater than ex , so the integral we set up to find the
area is:
Z 2
Z 2
2
(4x + 1) − (ex ) dx =
4x + 1 − ex dx = 2x2 + x − ex 1 =
1
1
2(2)2 + (2) − e2 − 2(1)2 + (1) − e1 = (10 − e2 ) − (3 − e) = 7 + e − e2
Example: Find the area bounded between the curves y = x2 − 4 and y = 4 − x2
Ans: We first look for the points of intersection of the two function by setting them equal
to each other:
x2 − 4 = 4 − x2 ⇒ 2x2 = 8 ⇒ x2 = 4 ⇒ x = ±2
y = 4 − x2
y = x2 − 4
Inside the interval [−2, 2], the function 4 − x2 is greater than the function x2 − 4, so we set
up the definite integral as:
Z
2
(4 − x2 ) − (x2 − 4) dx =
Z
−2
2
Z
2
2
2
4 − x2 − x2 + 4 dx =
− 2x2 + 8 dx = − x3 + 8x =
3
−2
−2
−2
2 3
2
16
16
3
− 16 =
− (2) + 8(2) − − (−2) + 8(−2) = − + 16 −
3
3
3
3
16
16
16
− + 16 −
+ 16 =
3
3
3
Example: Find the area bounded between x = 0, x = 2π, y = cos(2x), y = sin(x)
cos(2x)
sin(x)
Z
2π
|cos(2x) − sin(x)| dx
Ans: We are interested in
0
Inside the interval [0, 2π], for some values of x, cos(2x) > sin(x), and for some other values
of x, sin(x) > cos(2x). We set the two functions equal to each other to find the points of
intersection:
cos(2x) = sin(x)
1 − 2 sin2 (x) = sin(x)
2 sin2 x + sin(x) − 1 = 0
(2 sin(x) − 1)(sin(x) + 1) = 0
1
sin(x) = or sin(x) = −1
2
π
5π
3π
x=
or x =
or x =
6
6
2
π
5π
3π
∪
For x inside the intervals 0,
, 2π , cos(2x) > sin(x) (at x =
, the two functions
6
6
2
intersect, but the relationship between which is greater is not changed at that point.)
π 5π
For x inside the interval
,
, sin(x) > cos(2x)
6 6
The definite integral can be evaluated as:
Z
2π
|cos(2x) − sin(x)| dx =
A=
0
π/6
Z
Z
5π/6
cos(2x) − sin(x) dx +
0
Z
2π
sin(x) − cos(2x) dx +
π/6
cos(2x) − sin(x) dx
5π/6
π/6 5π/6 2π
sin(2x)
sin(2x)
sin(2x)
=
+ cos(x) + − cos(x) −
+
+
cos(x)
2
2
2
0
π/6
5π/6
sin(2(π/6))
sin(2(0))
=
+ cos(π/6) −
+ cos(0)
2
2
sin(2(5π/6))
sin(2(π/6))
+ − cos(5π/6) −
− − cos(π/6) −
2
2
sin(2(5π/6))
sin(2(2π))
+ cos(2π) −
+ cos(5π/6)
+
2
2
√
√
√ !
√ !
√
√ !
√
√ !
3
3
3
3
3
3
3
3
=
+
− (0 + 1) +
+
− −
−
+ (0 + 1) − −
−
4
2
2
4
2
4
4
2
!
√ !
√ !
√
3 3
3 3
3 3
−1 +
+ 1+
=
4
2
4
√
=3 3
Sometimes, the region is enclosed by two functions that are functions of y. In this case, it
may be easier to express the integral in terms of y instead of x to find the area:
Example: Find the area of the region bounded by x = −2y + 4 and x = y 2 + 1
x = −2y + 4
x = y2 + 1
Ans: While it is possible to find the area of the region by using definite integrals of functions
of x, it is more straight forward to set up a definite integral as function of y. Setting the
two curves equal to each other gives:
−2y + 4 = y 2 + 1
y 2 + 2y − 3 = 0
(y + 3)(y − 1) = 0
The points of intersection are at y = −3 and y = 1. Inside the interval −3 < y < 1, the
curve −2y + 4 always has a greater x value than the curve y 2 + 1, so we set up our integral
as:
Z
1
(−2y + 4) − (y 2 + 1) dy
A=
−3
Notice that our integral is in terms of y, not x.
Z
1
Z
1
Z
1
1
1
(−2y+4)−(y 2 +1) dy =
−2y+4−y 2 −1 dy =
−y 2 −2y+3 dy = − y 3 − y 2 + 3y 3
−3
−3
−3
−3
1
1
1
= − (1)3 − (1)2 + 3(1) − − (−3)3 − (−3)2 + 3(−3) = − − 1 + 3 − (9 − 9 − 9)
3
3
3
1
1 32
= − + 2 + 9 = 11 − =
3
3
3