# 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

### Example

### Solution

Draw y = x^{3}.

Draw y = x^{2} - x.

y = x^{2} - x = x(x - 1)

So the zeros of the graph are 0 and 1.

Quadratic Function: Find Zeros

Draw x = 2

between the curves.

Then color the bounded region.

If 0 ≤ x ≤ 2,

y = x^{3} is the upper function

and y = x^{2} - x is the lower funtion.

So the colored area is

∫_{0}^{2} [x^{3} - (x^{2} - x)] dx.

-(x^{2} - x) = -x^{2} + x

Solve the integral.

Definite Integral: How to Solve

The integral of x^{3} - x^{2} + x is

[1/4]x^{4} - [1/3]x^{3} + [1/2]x^{2}.

Integral of a Polynomial

Put 2 and 0

into [1/4]x^{4} - [1/3]x^{3} + [1/2]x^{2}.

[1/4]⋅2^{4} = [1/4]⋅16 = 4

-[1/3]⋅2^{3} = -8/3

+[1/2]⋅2^{2} = +2

-([1/4]⋅0^{4} - [1/3]⋅0^{3} + [1/2]⋅0^{2}) = -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

So 10/3 is the answer.

## Example 2

### Example

### Solution

Draw y = x^{2} - 3x.

y = x^{2} - 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 = x^{2} - 3x and y = x.

So set x^{2} - 3x = x.

Solve x^{2} - 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 = x^{2} - 3x is the lower funtion.

So the colored area is

∫_{0}^{4} [x - (x^{2} - 3x)] dx.

x - (x^{2} - 3x) = x - x^{2} + 3x

x + 3x = +4x

Solve the integral.

The integral of -x^{2} + 4x is

-[1/3]x^{3} + 4⋅[1/2]x^{2}.

+4⋅[1/2]x^{2} = +2x^{2}

Put 4 and 0

into -[1/3]x^{3} + 2x^{2}.

-[1/3]⋅4^{3} = -64/3

+2⋅4^{2} = +2⋅16

-(-[1/3]⋅0^{3} + 2⋅0^{2}) = -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

So 32/3 is the answer.