Math 201
Lia Vas
Areas between Curves
If f (x) and g(x) are two continuous functions defined on interval [a, b] such that f (x) ≥ g(x) for
all x ∈ [a, b] then the
area between the graphs of f and g on [a, b] is
A=
Z b
(f (x) − g(x)) dx.
a
The area between two intersecting curves f (x) and g(x) can be found following the steps below.
1. The bounds of integration are the intersections of the two curves and can be obtained by solving
f (x) = g(x) for x. Note that you can have more than two solutions and more than one relevant
region (e.g. problems 8 and 9 below).
2. If f and g intersect only at x = a and x = b, determine which function has larger values on
(a, b). In some cases, graph may be telling enough. If it is not, take an x-value, say x = c,
between a and b and plug it in both functions. Comparing f (c) and g(c) determines which
function is greater on (a, b). Be careful to pick a point within (a, b) i.e. between a and b.
3. If f (c) is larger than g(c),
then find the area as
R
then find the area as ab (g(x) − f (x)) dx.
Rb
a (f (x)
− g(x)) dx. If g(c) is larger than f (c),
If x = f (y) and x = g(y) and you need to find the area between them, you can interchange the
variables and find the area between y = f (x) and y = g(x).
If you need to find the area between three curves, graph the curves and divide the total area into
(at least) two regions, each of which is between a pair of curves.
Practice Problems. Find the area of the region between the given curves.
1. y = x2 + 3, y = x, x = −1, x = 1
2. y = 4x2 , y = x2 + 3
3. y = x2 , y = x
√
4. y = x + 3, y =
x+3
2
5. y = x2 , y 2 = x
6. y 2 = x, x − 2y = 3
7. x = y 3 − y, x = 1 − y 2
8. y = x3 , y = 3x2 − 2x
9. y = x3 , y = x
10. y = 3x − 3, y = 2 − 2x, y =
x
2
+2
11. y = x, y = 2x, y = 6 − x
2
+16x+9
12. A botanist knows that a certain species of oak tree grows at a rate of 4x 2x+4
feet per year,
where x is the age of the tree in years. When restricting the light, the oak tree grows at a rate
2x2 +12x+9
feet per year in x years. Determine how many fewer feet in growth will result from
2x+4
restricting the amount of light that tree receives when the tree is between 3 and 8 years old.
Solutions.
1. The bounds of integration are given to be -1 and 1. Using either the graph or plugging a point
from (-1,1) into both curves (for example 0), you can see that
y = x2 +3 is greater than y = x on
R
3
2
1
(-1,1) (3 = 02 +3 > 0). Thus, the area can be found as A = −1 (x2 +3−x)dx = x3 −3x− x2 = 20
.
3
2. Find the intersections first. 4x2 = x2 + 3 ⇒ 3x2 = 3 ⇒ x2 = 1 ⇒ x = 1 and x = −1. On
interval (−1, 1), the curve y = x2 + 3 is greater than
y = 4x2 (you can plug
0 to see that:
R1
R
1
2
2
2
2
(3 = 0 + 3 > 4(0) = 0). So, the area is A = −1 (x + 3 − 4x )dx = −1 (3 − 3x2 )dx =
3
3x − 3x3 |1−1 = 3 − 1 + 3 − 1 = 4.
3. Intersections: x2 = x ⇒ x2 − x = x(x − 1) = 0 ⇒ Rx = 0 and x = 1. On interval (0, 1), the
3
2
curve y = x is greater than y = x2 . The area is A = 01 (x − x2 )dx = x2 − x3 |10 = 21 − 13 = 16 .
√
2
4. Intersections: x + 3 = x+3
⇒ x+3 = (x+3)
⇒ 4(x+3) = (x+3)2 ⇒ 0 = (x+3)2 −4(x+3)0 =
2
4
(x +√
3)[x + 3 − 4] ⇒ 0 = (x + 3)(x − 1) ⇒ x = −3 andR x =
(−3, 1), the curve
√ 1. On interval
1
x+3
x+3
y = x + 3 is greater than y = 2 . The area is A = −3 ( x + 3 − 2 )dx. Use substitution
3/2
2
u = x + 3 to obtain that this integral is = 2(x+3)
− (x+3)
|1−3 = 16
− 4 = 34 = 1.33.
3
4
3
√
5. If y 2 = x, then y = ± x but√just the positive branch intersect the
curve y = x2 . Thus, you
√
can consider y = x2 and y = x. The curves intersect when x2 = x ⇒ x4 = x ⇒ x4 − x =
x(x3 − 1) = 0 ⇒ x = 0 and x3 = 1 ⇒ x = 0 and x = 1. The area can be computed as
R √
3/2
3
A = 01 ( x − x2 )dx = 2x3 − x3 |10 = 32 − 13 = 13 .
6. The line x − 2y = 3 intersect both positive and negative branch of y 2 = x ⇒ y = ±x.
Analyzing the graph, you can note that the area cannot be obtained using a single integral.
However, interchanging x and y variables, you obtain much easier set up: the region between
the curves x2 = y, y − 2x = 3 ⇒ y = x2 , y = 2x + 3, can be computed using single integral and
interchanging the variables does not impact the size of the intersecting region.
Find the intersections first. x2 = 2x + 3 ⇒ x2 − 2x − 3 = 0 ⇒ x = −1 and
x = 3. On interval
R3
2
(−1, 3), the curve y = 2x + 3 is greater than y = x . The area is A = −1 (2x + 3 − x2 )dx =
3
x2 + 3x − x3 |3−1 = 9 + 9 − 9 − (1 − 3 + 13 ) = 11 − 31 = 32
.
3
If you don’t notice that interchanging
the
creates an easier set up, you can compute
√ variables
R √
R √
)dx and obtain the same answer.
the area as A = A1 + A2 = 01 ( x − (− x)dx + 19 ( x − x−3
2
7. In this problem, the curves are given in terms of x. If you feel more comfortable working
with curves given in terms of y, you can interchange the variables and find the area between
y = x3 − x and y = 1 − x2 . Intersections: x3 − x + x2 − 1 = 0 ⇒ x(x2 − 1) + x2 − 1 = 0 ⇒
(x2 − 1)(x + 1) =R (x − 1)(x + 1)2 = 0 ⇒ x = −1 and x = 1. On (-1,1), y = 1 − x2 is greater.
3
4
2
1
The area is A = −1
(1 − x2 − x3 + x)dx = x − x3 − x4 + x2 |1−1 = 34 = 1.33.
8. Find the intersections. x3 = 3x2 − 2x ⇒ x3 − 3x2 + 2x = 0 ⇒ x(x2 − 3x + 2) = 0 ⇒
x(x − 1)(x − 2) = 0. So, the curves intersect at x = 0, x = 1, and x = 2. On interval [0,1], the
curve y = x3 is greater than y = 3x2 − 2x. On interval [1,2], the opposite is the case. Thus,
the total area can be obtained as the sum of the area over [0,1] and the area over [1,2].
A = A1 + A2 =
R1
0
(x3 − 3x2 + 2x)dx +
R2
1
(3x2 − 2x − x3 )dx. Compute this sum to be 21 .
9. Find the intersections. x3 = x ⇒ x3 − x = x(x2 − 1) = x(x − 1)(x + 1) = 0. So, the curves
intersect at x = 0, x = −1, and x = 1. On interval [-1,0], the curve y = x3 is greater than
y = x. On interval
[0,1], the opposite
is the case. Thus, the total area can be computed as
R0
R1
3
A = A1 + A2 = −1 (x − x)dx + 0 (x − x3 )dx. Compute this sum to be 21 .
10. Find the three intersections first.
(1) 3x − 3 = 2 − 2x ⇒ 5x = 5 ⇒ x = 1.
(2) 3x − 3 = x2 + 2 ⇒ 6x − 6 = x + 4 ⇒ 5x = 10 ⇒ x = 2.
(3) 2 − 2x = x2 + 2 ⇒ 4 − 4x = x + 4 ⇒ −5x = 0 ⇒ x = 0.
The relevant region consists of two parts: the area A1 on interval [0,1] between the upper curve
y = x2 + 2 and the lower y = 2 − 2x, and the area A2 on interval [1,2] between the upper curve
y = x2 + 2 and the lower y = 3x − 3.
A1 = 01 ( x2 + 2 − 2 + 2x)dx = 01 52 xdx = 54 x2 |10 = 54 = 1.25. A2 = 12 ( x2 + 2 − 3x + 3)dx = 12 (5 −
5
x)dx = 5x− 54 x2 |21 = 10−5−5+ 54 = 54 = 1.25. The total area is A = A1 +A2 = 54 + 54 = 52 = 2.5.
2
R
R
R
R
11. Find the three intersections first. (1) x = 6 − x ⇒ 2x = 6 ⇒ x = 3. (2) 2x = 6 − x ⇒ 3x =
6 ⇒ x = 2. (3) 2x = x ⇒ x = 0. The relevant region consists of two parts: the area A1 on
interval [0,2] between the upper curve y = 2x and the lower y = x, and the area A2 on interval
[2,3] between the upper curve y = 6 − x and the lower y = x.
Thus, the total area can be computed as A = A1 + A2 =
x2 2
| + (6x − x2 )|32 = 2 + (18 − 9 − 12 + 4) = 3.
2 0
R2
0
(2x − x)dx +
R3
2
(6 − x − x)dx =
12. The difference in the growth of the oak thee in the two different set up from 3 to 8 years
can beRfound as the difference
of the two definite integrals computing the total growth in two
R 8 2x2 +12x+9
8 4x2 +16x+9
cases: 3 2x+4 dx − 3 2x+4 dx. Note that it is easier to combine the two integrals and
R 2 +16x+9
R
2 +12x+9
2
2 −12x−9
then evaluate the resulting integral 38 4x 2x+4
− 2x 2x+4
dx = 38 4x +16x+9−2x
dx =
2x+4
R 8 2x2 +4x
3
2x+4
dx =
R 8 x(2x+4)
3
2x+4
dx =
R8
3
xdx =
x2 8
|
2 3
=
55
2
= 27.5 feet.