# Area under the Graph of a Function

How to find the area under the graph of a function: formula, proof, examples, and their solutions.

## Formula

If *y* = *f*(*x*) is above the *x*-axis, (in [*c*, *b*])

then the area is the integral of *f*(*x*).

But if *y* = *f*(*x*) is below the *x*-axis, (in [*a*, *c*])

then the area is the integral of -*f*(*x*).

So the area under *y* = *f*(*x*)

is the integral of |*f*(*x*)|.

Definite integration of absolute value functions

## Proof

If *y* = *f*(*x*) is below the *x*-axis, (in [*a*, *c*])*f*(*x*) is (-).

So the area of the sliced rectangle, *f*(*x*) *dx*, is (-).

[ (-)⋅(+) = (-) ]

So the integral of *f*(*x*) *dx* is (-).

But the sign of the area should be (+).

So, in order to change this (-) sign,

change the integrand *f*(*x*) into -*f*(*x*).

So the area below the *x*-axis is ∫_{a}^{c} -*f*(*x*) *dx*.

If *y* = *f*(*x*) is above the *x*-axis, (in [*c*, *b*])

the area above the *x*-axis is ∫_{c}^{b} *f*(*x*) *dx*.

Because of the change of the integrands, [-*f*(*x*), *f*(*x*)]

the integrand is always (+).

So the absolute value sign can be used

to express the area under the graph of *y* = *f*(*x*).

## Example 1

First find the bounded region.*y* = *x*^{2} - 2*x*

= *x*(*x* - 2)

So the zeros of *y* = *x*^{2} - 2*x* are 0 and 2.

Quadratic functions - factored form

In [0, 2],*y* = *x*^{2} - 2*x* is below the *x*-axis.

So the area in [0, 2] is

∫_{0}^{2} -(*x*^{2} - 2*x*) *dx*.

In [2, 3],*y* = *x*^{2} - 2*x* is above the *x*-axis.

So the area in [2, 3] is

∫_{2}^{3} (*x*^{2} - 2*x*) *dx*.

Solve the integrals.

Definite integration of polynomials

To make the denominators the same,

multiply 3/3 by 8.

## Example 2

First find the bounded region.

Graphing sine functions

In [0, *π*],*y* = sin *x* is above the *x*-axis.

So the area in [0, *π*] is

∫_{0}^{π} (sin *x*) *dx*.

In [*π*, 2*π*],*y* = sin *x* is below the *x*-axis.

So the area in [*π*, 2*π*] is

∫_{π}^{2π} (-sin *x*) *dx*.