# Area between Curves

How to find the area between two curves: formula, 2 examples, and their solutions.

## Formula

### Formula

The area between two curves is
the integral of the difference of the curves:
|f(x) - g(x)|.

y = f(x) is the upper curve function.
y = g(x) is the lower curve function.

Area under a Curve

## Example 1

### Solution

Draw y = x3.

Draw y = x2 - x.
y = x2 - x = x(x - 1)
So the zeros of the graph are 0 and 1.

Draw x = 2
between the curves.

Then color the bounded region.

If 0 ≤ x ≤ 2,
y = x3 is the upper function
and y = x2 - x is the lower funtion.

So the colored area is
02 [x3 - (x2 - x)] dx.

-(x2 - x) = -x2 + x

Solve the integral.

Definite Integral: How to Solve

The integral of x3 - x2 + x is
[1/4]x4 - [1/3]x3 + [1/2]x2.

Integral of a Polynomial

Put 2 and 0
into [1/4]x4 - [1/3]x3 + [1/2]x2.

[1/4]⋅24 = [1/4]⋅16 = 4
-[1/3]⋅23 = -8/3
+[1/2]⋅22 = +2

-([1/4]⋅04 - [1/3]⋅03 + [1/2]⋅02) = -0

4 + 2 = 6

To solve 6 - 8/3,
multiply 3/3 to 6.

Then 6⋅3/3 - 8/3 = 18/3 - 8/3.

18/3 - 8/3 = 10/3

## Example 2

### Solution

Draw y = x2 - 3x.
y = x2 - 3x = x(x - 3)
So the zeros of the graph are 0 and 3.

Draw y = x.

Then color the region
between the graphs.

The colored region is
from x = 0 to x = ?.

To find the value of ?,
find the intersecting points of y = x2 - 3x and y = x.

So set x2 - 3x = x.

Solve x2 - 3x = x.

Then x = 0, 4.

x = 0, 4

See the graphs.
The intersecting points are
x = 0 and x = ?.

So ? = 4.

If 0 ≤ x ≤ 4,
y = x is the upper function
and y = x2 - 3x is the lower funtion.

So the colored area is
04 [x - (x2 - 3x)] dx.

x - (x2 - 3x) = x - x2 + 3x

x + 3x = +4x

Solve the integral.

The integral of -x2 + 4x is
-[1/3]x3 + 4⋅[1/2]x2.

+4⋅[1/2]x2 = +2x2

Put 4 and 0
into -[1/3]x3 + 2x2.

-[1/3]⋅43 = -64/3
+2⋅42 = +2⋅16

-(-[1/3]⋅03 + 2⋅02) = -0

+2⋅16 = 32

-64/3 = -(32⋅2)/3

32 - (32⋅2)/3 = 32(1 - 2/3)

Common Monomial Factor

1 - 2/3 = 1/3

32⋅[1/3] = 32/3