# Area under a Curve

How to find the area under a curve (between the curve and the x-axis): formula, 2 examples, and their solutions.

## Formula

The area under a curve

(between the curve and the x-axis)

is the integral of the curve function.

Riemann Sum to Definite Integral

If the curve is above the x-axis,

the area is the integral of f(x):

∫_{a}^{c} f(x) dx.

But if the curve is below the x-axis,

the area is the integral of -f(x):

∫_{c}^{b} [-f(x)] dx.

(The minus sign is added

to make the sign of the integral plus.)

So the area under a curve

is the integral of |f(x)|:

∫_{a}^{b} |f(x)| dx.

## Exampley = x^{2} - 2x, x = 3, x: 0 ~ 3

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

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

So the zeros of the graph are 0 and 2.

Quadratic Function: Find Zeros

Draw x = 3.

Then color the region

that is bounded by these two graphs and the x-axis.

If 0 ≤ x ≤ 2,

the area is below the x-axis.

So the below area is

the integral of -f(x):

∫_{0}^{2} -(x^{2} - 2x) dx.

If 2 ≤ x ≤ 3,

the area is above the x-axis.

So the above area is

the integral of f(x):

∫_{2}^{3} (x^{2} - 2x) dx.

So this is the area of the bounded regions.

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

Solve the integral.

Definite Integral: How to Solve

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

2⋅[1/2]x^{2} - [1/3]x^{3}.

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

[1/3]x^{3} - 2⋅[1/2]x^{2}.

Integral of a Polynomial

2⋅[1/2]x^{2} - [1/3]x^{3}

= x^{2} - [1/3]x^{3}

[1/3]x^{3} - 2⋅[1/2]x^{2}

= [1/3]x^{3} - x^{2}

Put 2 and 0

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

Put 3 and 2

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

2^{2} = 4

-[1/3]⋅2^{3} = -[1/3]⋅8

-(0^{2} - [1/3]⋅0^{3}) = -0

+[1/3]⋅3^{3} = +9

-3^{2} = -9

+[1/3]⋅2^{3} = +[1/3]⋅8

-2^{2} = -4

-[1/3]⋅8 = -8/3

Cancel +9 and -9.

-([1/3]⋅8 - 4) = -8/3 + 4

4 + 4 = 8

-8/3 - 8/3 = -16/3

To solve 8 - 16/3,

multiply 3/3 to 8.

Then 8⋅3/3 - 16/3 = 24/3 - 16/3.

24/3 - 16/3 = 8/3

So 8/3 is the answer.

## Exampley = sin x (0 ≤ x ≤ 2π)

y = sin x (0 ≤ x ≤ 2π) means

one cycle of y = sin x.

So draw a cycle of y = sin x.

Sine: Graph

Color the region

that is bounded by y = sin x and the x-axis.

If 0 ≤ x ≤ π,

the area is above the x-axis.

So the above area is

the integral of f(x):

∫_{0}^{π} sin x dx.

If π ≤ x ≤ 2π,

the area is below the x-axis.

So the below area is

the integral of -f(x):

∫_{π}^{2π} (-sin x) dx.

So this is the area of the bounded regions.

+∫_{π}^{2π} (-sin x) dx = -∫_{π}^{2π} sin x dx

Solve the integral.

The integral of sin x is

-cos x.

-[-cos x]_{π}^{2π} = +[cos x]_{π}^{2π}

Put π and 0

into -cos x.

Put 2π and π

into cos x.

-cos π = -(-1)

-(-cos 0) = -(-1)

+cos 2π = +1

-cos π = -(-1)

Cosine Values of Commonly Used Angles

-(-1) = +1

+1 + 1 + 1 + 1 = 4

So 4 is the answer.