# Area between curves ```Area between curves
Vertical slices.
The area between the curves y = f(x) and y = g(x) over a 6 x 6 b is obtained as follows.
Zb
A=
(f(x) − g(x)) dx
a
Observe that the rectangular slice has length given by f(x) − g(x) and width dx.
y
y = f(x)
y = g(x)
a
b
x
dx
Horizontal slices.
The area between the curves x = f(y) and x = g(y) over c 6 y 6 d is obtained as follows.
Zd
A=
(f(y) − g(y)) dy
c
Observe that the rectangular slice has length given by f(y) − g(y) and width dy.
y
x = g(y)
x = f(y)
d
dy
x
c
Gilles Cazelais. Typeset with LATEX on January 13, 2007.
Example 1. Find the area bounded by the two parabolas y = x2 and y = 2x − x2 .
Solution. We first find the points of intersection of the parabolas by solving
x2 = 2x − x2 ⇐⇒ 2x2 − 2x = 0 ⇐⇒ 2x(x − 1) = 0.
The solutions are x = 0 or 1. The points of intersections are then (0, 0) and (1, 1).
y
y = 2x − x2
1
y = x2
0
x
1
We find the area using vertical slices as follows.
Z1
A=
(2x − x2 − x2 ) dx =
0
Z1
(2x − 2x2 ) dx
0
2x3 1
= x2 −
3 0
=1−
1
2
=
3
3
Example 2. Find the area between the line y = x − 1 and the parabola y2 = 2x + 6.
Solution. By solving y + 1 = (y2 − 6)/2, we can deduce that the points of intersection are
(−1, −2) and (5, 4).
y
y2 − 6
x=
2
(5, 4)
4
x=y+1
x
(−1, −2)
−2
We find the area using horizontal slices as follows.
2
Z4 y −6
A=
(y + 1) −
dy
2
−2
4
Z4
1 y3
y2
1 2
=
− 2 y + y + 4 dy = −
+
+ 4y 2 3
2
−2
−2
1
8
= − 6 (64) + 8 + 16 − 6 + 2 − 8 = 18
```